Does a closed curve exist for which a square cannot intersect it 8 or more times?

I don't have an answer for the most general case, but I have solved several of the simpler cases. As a summary of the below results, we have solved all the cases where the curve is simple and any one of: convex, or differentiable, or not BV, or not semi-differentiable (in a precise sense). In all of these cases, Maxime will win.

After working on it for a while, I think completing the problem in full will be extremely difficult, and require much more research. Rather than continue to add on edits, I'm just going to leave this half answer here for the future, in case anyone gets some use out of it. Feel free to adapt any ideas here into your own answer, if you can figure out the general version.

Let $\sigma:X\rightarrow \mathbb{R}^2$ continuously draw a positively oriented closed loop, and suppose that Maxime cannot win on $\sigma$. We may varyingly assume $X=S^1$ or $X=[0,1]$ as needed. Since $S^1$ is compact and connected, and $\sigma$ is continuous, then its image is also compact and connected. We assume that $\sigma$ is simple; the non-simple cases are likely much easier for Maxime.

$\textbf{Claim}$: There is no single line/ray/segment which $\sigma$ intersects more than $8$ times.

proof: Put a square taking that line as an edge, Maxime wins. $\square$

Suppose $\delta:S^1\rightarrow \mathbb{R}^2$ is a Jordan curve, then by the Jordan Curve theorem, $\delta$ partitions the plane into an interior and exterior region, where $\delta$ is the boundary. Given some point $t\in S^1$ and injective continuous curve $\gamma:X\rightarrow \mathbb{R}^2$ intersecting at $\delta(t)$, we say that $\gamma$ is interior tangent to $\delta$ at $t$ when there is a neighborhood $\delta(t)\in U\subseteq\mathbb{R}^2$ over which $\gamma$ stays in the interior region defined by $\delta$ (except at $\delta(t)$). We similarly define exterior tangents. If they intersect but not in the previous cases, then we say $\gamma$ crosses $\delta$ at $t$. Naturally these cases are all mutually exclusive. Notice that if both $\delta,\gamma$ are Jordan curves, this induces at least five different sorts of intersection schemes.

$\textbf{Claim}$: No ray crosses $\sigma$ at its endpoint.

proof: By contradiction, suppose we had a ray $r:(-\infty,0]\rightarrow \mathbb{R}^2$ which did cross $\sigma$ at some $\sigma(t)=r(0)$, then for every neighborhood $U\subseteq \sigma(t)$, the image of $r$ has some points not in the interior and some not in the exterior of the curve $\sigma$, so we may construct a strictly increasing series $x_n\rightarrow 0$ over which $r(x_n)$ alternates from not interior to not exterior. We can then construct a series of intermediaries $y_n\rightarrow 0$ for which $r(y_n)$ is on the boundary, that is, on $\sigma$. By extending $r$ to a line, we have infinitely many intersections, Maxime wins. $\square$

$\textbf{Claim}$: The curve $\sigma$ is semi-differentiable, in that the following directional limits converge: $$\sigma'_+(t_0)=\lim_{t\rightarrow t_0^+}\frac{\sigma(t)-\sigma(t_0)}{||\sigma(t)-\sigma(t_0)||} \hspace{12mm} \sigma'_-(t_0)=\lim_{t\rightarrow t_0^-}\frac{\sigma(t_0)-\sigma(t)}{||\sigma(t)-\sigma(t_0)||}$$

proof: We shall prove that the right ($-$) limit exists, the other case is similar. Let $v(t_0,t)=\frac{\sigma(t_0)-\sigma(t)}{||\sigma(t)-\sigma(t_0)||}$ as in the above limit. This is always a unit vector, and is always defined for $t\neq t_0$ since $\sigma$ has no self intersections. Moreover, it is continuous by $t$. Since $v(t_0,t)$ is a unit vector, then it lies on the circle $S^1$. Notice that if $v(t_0,t)=u$, then we can construct a ray ending at $\sigma(t_0)$ with orientation $u$ passing through $\sigma(t)$. Since this ray doesn't cross $\sigma$, then $v(t_0,t)$ can take the value $u$ only finitely many times. Since $S^1$ is sequentially compact, then as $t\rightarrow t_0^+$, there is at least one limit point for $v(t_0,t)\rightarrow v_1$. By contradiction, suppose we had any second limit point $v(t_0,t)\rightarrow v_2$, then $v_1,v_2$ splits $S^1$ into two arcs. Select points $u_1,u_2$ in different arcs, and notice that now $u_1,u_2$ also split $S^1$ into two arcs, separating $v_1,v_2$. By the above, $v(t_0,t)$ takes on the values $u_1,u_2$ only finitely many times. By continuity, after the last time it takes on either value, $v(t_0,t)$ must stay within a single arc, say that of $v_1$. But now the other arc is a neighborhood for $v_2$ which $v(t_0,t)$ never enters again; contradiction. Thus $v_2$ does not exist, and the limit $v_1$ is unique. $\square$

Note that these directional derivatives are not quite the standard ones, usually we would be dividing by $t_0-t$ instead. This is why we had to negate the expression in the left limit.

$\textbf{Claim}:$ The curve $\sigma$ is of bounded variation.

Firstly, we can prove that the variation of every projection is finite. More simply, for each unit vector $v_0\in S^1$, there is some bound $B$ such that every partition $[t_1,t_2),[t_2,t_3),..[t_n,t_{n+1}]$ of the interval $[0,1]$ into subintervals obeys the following estimate: $$B>\sum_{k=1}^{n} |(\sigma(t_{k+1})-\sigma(t_{k}))\cdot v_0|$$

This follows from a sort of continuous pigeon hole principle, which I'll state as an unproven lemma.

$\textbf{Lemma}$: let $I\subseteq \mathbb{R}$ be a closed interval with length $d$, and suppose we have a series of open intervals $I_1,...,I_n$ all with $I_k\subseteq I$, and such that the sums of their lengths $I_1+...+I_n$ is $L$. If it is the case that $L>nd$ for some integer $n$, then there is a point $x\in I$ which is also in at least $n$ many of the sub intervals.

For our purposes, $n=8$, we may take $d$ to be the diameter of the curve $\sigma$, we may take $I$ to be the projection of $\sigma\cdot v_0$ (so that the parallel extension of $v_0$ is identified with $\mathbb{R}$), and likewise all the $I_n$ to be the projections of the images $\sigma((t_{n},t_{n+1}))\cdot v_0$ which have lengths $|(\sigma(t_{k+1})-\sigma(t_{k}))\cdot v_0|$. If our conjecture holds, then the variation of our projection must be bounded by $B=8d$. If it weren't, then we can find our point $x$, and select our line as the pre-projection of that point, and this will intersect at least $n$ many of the arcs $\sigma((t_{n},t_{n+1}))$, which correlates to $n$ intersections of the line through $\sigma$ (since our sub-intervals are disjoint). Since $n=8$, this violates our assumptions, so the variation of the projection is bounded by $B=8d$.

By combining the above estimate for two orthogonal directions, we can prove that the total variation acquired by $\sigma$ across the $t_n$ series is also bounded, instead by $16d$. In particular, letting $v_1$ be a unit vector orthogonal to $v_0$, we get: $$|\sigma(t_{n+1})-\sigma(t_{n})| \leq |(\sigma(t_{n+1})-\sigma(t_{n}))\cdot v_0|+|(\sigma(t_{n+1})-\sigma(t_{n}))\cdot v_1|$$ $$\sum_{n=1}^{\infty} |\sigma(t_{n+1})-\sigma(t_{n})|\leq 8d+8d=16d$$

Since $d$ was independent of the $t_n$ partition, this proves that $\sigma$ is of bounded variation. $\square$

If needed, we may re-parametrize $\sigma$ as needed to assume it is rectified.

$\textbf{Lemma}:$ If $\sigma$ is interior tangent to a rectangle $R$ on some edge $E$, then for any neighborhood $R(p)\in U\subseteq\mathbb{R}^2$ there is some $r>0$ such that: a contraction of $R$ by pulling in the edge $E$ (leaving its parallel fixed) of any positive amount less than $r$ causes $\sigma$ to cross the new rectangle twice within $U$ on the new edge.

proof: For simplicity, assume the edge $E$ is the left vertical edge of $R$, and parametrize $\sigma$ so that $t_0\in(0,1)$ has our intersection $\sigma(t_0)=R(p)$ on the edge. From tangency, find a neighborhood $\sigma(t_0)\in U_1\subseteq U$ over which $\sigma$ stays interior to $R$ except at the intersection $R(p)$. Let $t\in V\subseteq \sigma^{-1}(U_1)$ be open and connected. Let $t_-<t_0<t_+$ be points in $V$, and let $x_0$ be the minimum of the projections $\pi_x(\sigma(t_\pm))$ to the $x$-axis. We may now take $r$ to be the positive distance between $x_0$ and $\pi_x(E)$. We now have the projection $\pi_x(\sigma(t_\pm))\geq x_0>x_0-r$, while $\pi_x(\sigma(t_0))=\pi_x(R(p))=\pi_x(E)=x_0-r$. By IVT, we must have $\pi_x(\sigma(t))$ take on every value between $x_0$ and $x_0-r$ between $t_-$ and $t_0$, and then again from $t_0$ to $t_+$. Pulling in the edge $E$ by any amount in this range thus induces two unique intersections. $\square$

The above proof can be repeated for exterior tangents, by extending the edge instead. We can also repeat the proof in the case our intersection is on a corner, to contract both edges simultaneously (leaving the opposite corner fixed). We can repeat a slight variation of the proof to show that if the point is a crossing instead of a tangent, that a sufficiently small adjustment will keep one intersection nearby.

$\textbf{Claim}$: If there is a square to which $\sigma$ is interior tangent at $n$ points, and crossing at $k$ other points, then there is a square to which $\sigma$ is crosses at $2n+k$ points.

Let $\epsilon>0$ be less than a sixteenth the distance between any two intersections. Find $r$ small enough that contracting every edge equally by $r$ keeps an intersection within an $\epsilon$ ball of the old ones, and such that every tangent splits into two intersections within an $\epsilon$ ball of the old tangent. Contract all four sides equally by this amount. This creates a new square with $2$ new crossings around each old tangent, and $1$ crossing near each old crossing. By the conditions on $\epsilon$, these must all be distinct intersections, giving our new square at least $2n+k$ crossings. $\square$

This also works for exterior tangents, by expanding. This shows that tangents are essentially worth two crossings, so long as the tangents are all of the same type. Notice that the tangents and crossings really do behave differently though. This example has a square with four crossings, but no infinitesimal shrinking or expanding will turn any one of them into two.

$\textbf{Claim}$: The curve $\sigma$ is not convex.

proof: Suppose it were, then the curve is also locally monotone. This is a solved case of the inscribed square problem. Any Inscribed square to a convex curve must have $\sigma$ exterior tangent at all four corners. Expand for eight. Maxime wins. $\square$

Note that this implies there is some least bounding rectangle with two contacts on a single edge.

$\textbf{Claim}$: If there is a rectangle to which $\sigma$ is interior tangent at $n$ points and crossing at $k$ points, across three of its shorter edges, then there is a square to which $\sigma$ crosses $2n+k$ times.

proof: Extend the remaining fourth edge into a square, which has all the same crossings and tangents. Expand for $2n+k$ crossings. $\square$

So the least bounding rectangle just mentioned either has the double contact on a short edge, or has one of them in a corner, since otherwise the adjacent short edges would both get their own tangents.

$\textbf{Claim}:$ There is no rectangle $R$ such that $\sigma$ is interior tangent to $R$ at two of its adjacent corners.

Let $E$ be the corresponding edge of the rectangle $R$. Perform a contraction of the edge $E$ to $E'$ small enough to induce four distinct crossings over $E'$, and such that the distance between $E,E'$ is shorter than their length. Construct a rectangle taking $E,E'$ as edges, so that the original corners of $R$ are preserved. Notice that this new rectangle has four crossings on $E'$, and two tangents on the short perpendiculars (at the corners). Maxime wins. $\square$

If needed, here is a visual aid for the previous proof.

$\textbf{Claim}$: There is a square which crosses $\sigma$ six times.

proof: This proof is an adaption of the method used here by Marco. Find $\theta_\infty$ for which the least bounding rectangle $R_{\theta_\infty}$ has the longest possible side length. There must be contacts on the shorter edges, and they can't both be in corners: if they were in opposite corners, we could rotate to make a longer rectangle, otherwise they would be in adjacent corners, which we can't have. There must also be at least one point on a long edge, since our rectangle is least bounding. We now have three interior tangents among three of the shorter sides of a rectangle, which we can extend to give us a square with three interior tangents. Contract for six crossings. $\square$

$\textbf{Claim}$: There is a square for which $\sigma$ is interior tangent on all four edges.

For each angle $\theta$, let $R_\theta$ be the least bounding rectangle oriented parallel to the angle $\theta$ from standard reference. Notice that $\sigma$ must be interior tangent to $R_\theta$ on all four edges. Let $d(\theta)$ be the length of $R_\theta$ in the direction of $\theta$, that is, $d(\theta)$ is the diameter of $\sigma$ when measured in the direction $\theta$. Notice that $d(\theta)$ is both continuous and $\pi$ periodic. By intermediate value theorem, it follows that there must be some angle $\theta_0$ such that $d(\theta_0)=d(\theta_0+\frac{\pi}{2})$. When this is satisfied, the rectangle $R_{\theta_0}$ is in fact a square, and satisfies our claim. $\square$

Notice that this does not necessarily imply our square is tangent on four distinct points. In particular, this figure shows a least bounding square with only three points of contact. This is only possible if one of the tangents is at a corner however, at which $\sigma$ takes a sharp turn. Label this point $c_0$ on $\sigma$, and note that every single rectangle $R_\theta$ has this point somewhere on its edge.

$\textbf{Claim}$: For each of the least bounding rectangles $R_\theta$, the tangents on the long edges which are not at corners must have their horizontal separation (along that edge) not less than the length of the short edges of $R_\theta$.

proof: Suppose not, then $R_\theta$ has $p_1,p_2$ on the long edges (not corners) with horizontal distance less than the short length of $R_\theta$. Pull in the short edges until we reach a square, while keeping $p_1,p_2$ on the edge and not in a corner of the new square. Notice that $p_1,p_2$ are still interior tangents to our new square. Moreover, since $R_\theta$ was the least bounding rectangle, it had contacts on both short edges, which are now exterior to our square, and thus must form two new crossings each. Our square now has two interior tangents and four crossings. Maxime wins. $\square$

This isn't an answer, but I need to post a picture in order to demonstrate that the crescent idea doesn't work. See the following:

This handles the case where the upper arc subtends more than $\pi/2$ radians. And in the "sufficiently flat" case, we have:

which I think is quite interesting since one side has zero intersections but the opposite side makes up for this by having four. I think this is a key configuration that demonstrates that any non-convex shape admits a line that intersects the boundary at four distinct points.

Here is my current proposed example, I do not know if it works, did not verify it (it doesn't look like it does), but it is what I came up with after my "crescent shape" idea was proved wrong. It seems to me that the convex case is easy to deal with, and as for the non-convex case, one could take a supporting line touching the curve in (generally) just two points (and so that the curve remains in only one of the two half-planes determined by the line). Moving this supporting line a bit, one gets 4 intersections, and may then consider more cases, some of which I know how to deal with, and the following one in which I do not seem to know what to do (yet).

It looks to me that one could deal with the above shape in a way similar to the convex case, ignoring the long tail, taking a suitable inscribed square in the"oval" part of the shape (at the right of the above picture) and enlarging this square a bit to get 8 intersections.

The following idea (the first version of my answer) doesn't work, as indicated in a comment by heropup, and explained in detail in the answers by Jade Vanadium, and by heropup.

Take a half moon (crescent), as shown on the picture.
There are at most 6 intersections with any square (even if 8 intersections with suitable rectangles). The crescent needs to be rather flat, and not like the following one.