Is the complement of countably many disjoint closed disks path connected?

Let $\{D_n\}_{n=1}^\infty$ be a family of pairwise disjoint closed disks in $\mathbb{R}^2$. Is the complement $$ \mathbb{R}^2 -\bigcup_{n=1}^\infty D_n $$ always path connected?

Here “disk” means a round, geometric disk. (As shown below, the answer is no if we allow arbitrary topological disks.)

Note that the union $\bigcup_{n=1}^\infty D_n$ can be dense in the plane. For example, it's possible to find a collection of disjoint closed disks of positive radius whose union contains $\mathbb{Q}\times\mathbb{Q}$.

It is easy to show that the complement of a countable set of points in the plane is always path connected. In particular, if $S \subset\mathbb{R}^2$ is countable, then there is always path between any two points in $\mathbb{R}^2-S$ consisting of two line segments.

It is not true that the complement of a countable set of disjoint line segments is path connected, as shown in the following figure.

By thickening the line segments slightly, one can find a countable collection of disjoint topological disks whose complement is not path connected.

Contrary to the request of the OP, this is still a sketch. I put it here because it points to some relevant literature, hoping that somebody more expert than me will help to clarify it. The question reminded me of this paper by Fischer and Zastrow: the relevant part is the end of p. 12. See also this answer by George Lowther.

You can assume that $\cup D_n$ is dense. Now choose some nice diffeomorphism $\phi:\mathbb{R}^2\to B_1(0)$. Once you apply it, the images of the disks are no longer disks, but $X:=\overline{B_1(0)}\setminus\bigcup_n \phi(\text{int }D_n)$ is still a planar 1-dimensional Peano continuum without local cutpoints (I'm not sure about the meaning of 1-dimensional..), so it is homeomorphic to the Sierpinski carpet by a theorem of Whyburn (Whyburn, Topological characterization of the Sierpinski curve, Fund. Math. 45 (1958) 320–324).

Our space is homeomorphic to $X\setminus \left(\partial B_1(0)\cup\bigcup_n\phi(\partial D_n)\right)$, but these sets we are removing are characterized as being the only non-separating simple closed curves. So under the homeomorphism with the Sierpinski carpet they map exactly to the boundaries of the holes (plus its exterior boundary) and we are left to show that:

The Sierpinski carpet is still path connected even after removing the boundaries of its holes and its exterior boundary.

But this is easy. Call $S'$ this new awful noncompact carpet. Put $C_1:=[\frac{1}{3},\frac{2}{3}]$, $C_2:=[\frac{1}{9},\frac{2}{9}]\cup [\frac{4}{9},\frac{5}{9}]\cup [\frac{7}{9},\frac{8}{9}]$, and so on. We have $S'=(0,1)^2\setminus\bigcup_{i=1}^\infty C_i\times C_i$.
Put also $G:=(0,1)\setminus\bigcup_{i=1}^\infty C_i$ and observe that $T:=(0,1)\times G\cup G\times (0,1)$ is path connected and $T\subset S'$.
Now fix any $p_0\in T$. Given any $x\in S'$, we wish to connect it to $p_0$ with a path in $S'$. Let $x\in Q_1$ where $Q_1$ is one of the $9$ closed cubes with side $\frac{1}{3}$ (those in which $[0,1]^2$ is divided when constructing the Sierpinski carpet). Choose some $p_1\in T\cap \text{int }Q_1$ and connect $p$ to $p_1$ with a path of length $<3$ (namely a path formed by at most three line segments).
Replace $[0,1]^2$ with $Q_1$ and iterate this step using self-similarity. Iterating infinitely many times we get a sequence $p_n\to x$ and paths in $S'$ connecting $p_n$ to $p_{n+1}$ with length $<\frac{3}{3^n}$, so that concatenating them we get a path from $p_0$ to $x$. $\blacksquare$

I think it is path-connected.

Let $S = \{ D_n \, | \, n \in \mathbb{N} \}$ be the set of closed disjoint disks under consideration. For any integer $m \ge 1$, let $N_m = \{ n \in \mathbb{N} \, | \, \mathrm{radius}(D_n) \in [1/m, 1/m-1) \}$ (with the convention $1/0 = \infty$) and $S_m = \{ D_n \, | \, n \in N_m \}$.

Let $P, Q \in \mathbb{R}^2 \backslash S$ and consider some path $\gamma_0$ between the two, say the straight line segment. For area considerations, there is a finite number of disks in $S_1$ intersecting $\gamma_0$ ; Let $J_1$ denote the reunion of these disks and $K_1$ denote the reunion of $J_1$ and $\gamma_0$. For area and compactness considerations, there is an $\epsilon_1 > 0$ such that the closed $\epsilon_1$-neighborhood $K'_1$ of $K_1$ intersects no other disks of $S_1$. For similar reasons, there is an $\delta_1 \in (0, \epsilon_1/4)$ such that the open $\delta_1$-neighborhood $J'_1$ of $J_1$ only intersects disks of $S$ that are contained in the open $\frac{\epsilon_1}{2}$-neighborhood of $J_1$ (and which have as such rather little radius). Consider the compact and path-connected set $R_1 := K'_1 \backslash J'_1$. One can perturb $\gamma_0$ to a continuous path $\gamma_1$ in $R_1$. Note that this implies that $\gamma_1$ is at a distance $\delta_1 > 0$ of any disk in $S_1$.

It is important to note why $J_1'$ was chosen this way : $\delta_1$ is small enough compared to $\epsilon_1$ to assure that $R_1 \backslash S_2$ is also path-connected. Moreover, points that have been pushed out at the first step don't have to be pushed out a lot in order to avoid disks of $S_2$, since it is sufficient (near those points) to perturb the path inside the intersection of $R_1$ and the $\frac{\epsilon_1}{2}$-neighbordhood of $J_1$. Points that have not been pushed out at the first step might be pushed a lot at the second step, but the same reasoning shows that they won't have to be perturbed a lot from then on.

This suggests that a by careful induction process, we can construct a nested sequence of compact path-connected sets $R_1 \supset R_2 \supset R_3 \supset \dots$ and a sequence of paths $\gamma_m \subset R_m$ joining $P$ and $Q$ such that $d(\gamma_j, S_m) > \delta_m > 0$ whenever $j \ge m$. If the perturbation process is done right (that is, if it is kind of minimal), the sequence $\gamma_m$ is equicontinuous. By the Arzelà-Ascoli theorem, we deduce that the set $\{ \gamma_m \, | \, m \in \mathbb{N}\} \subset C(I, \mathbb{R}^2)$ is relatively compact. Hence, there exists a continuous path $\gamma : I \to \mathbb{R}^2$ (still joining $P$ and $Q$) which is arbitrary well approximated by the set of $\gamma_m$. In fact, since the $R_m$s form a decreasing sequence of nested compact sets, we deduce that $\gamma \subset \cap_{m \in \mathbb{N}} R_m$. By construction, we see that $d(\gamma, S_m) > \delta_m > 0$ and consequently, $\gamma$ intersects none of the disks $D_n$.

So here is my attempt to clean up and clarify the other answer. We pick $A,B \in \Bbb R^2 \setminus S$

For some disks $D_n = B(c_n,r_n)$ we're going to find a "thickening" of $D_n$, $T_n$, such that its border $B_n$ is disjoint from every disk, and find a continuous projecti

on $f_n : T_n \setminus \{ c_n\} \to B_n$. We also want the $T_n$ to be disjoint form each other, to not contain $A$ and $B$, and we want every disk to be included in exactly one of them.

By area considerations, for any compact set $K$ and radius $r$, there is a distance $d(K,r)$ such that for any disk $D$ of radius larger than $r$, the distance from $K$ to $D$ is either $0$ or strictly larger than $d(K,r)$.
Also I denote $B(K,d)$ the set of points that are at distance less than or equal to $d$ from $K$

We start by enumerating the disks, and repeatedly look at the smallest index $n$ of a disk not contained in a already constructed thickening of a previous disk. We have to build a thickening $T_n$ of $D_n$.

We start with $K_0 = D_n$ and $\phi_0 : K_0 \setminus \{c_n\} \to K_0$ to be the radial projection onto its boundary.

Because there have been finitely many thickenings constructed so far, $K_0$ is at a positive distance from their reunion, and so there is a distance $d_0$ such that $B(K_0,d_0)$ doesn't intersect any thickening, any disk of radius $>1$, nor $A$ or $B$. We will want to have $T_n$ to stay within distance $d_0$ of $K_0$.
Let $e_0 \le \min(d_0/6,d(K_0,d_0/6))$ such that $L_0 = B(K_0,e_0)$ is not tangent to any disk (this is possible because there are countably many disks and uncountably many reals). There is an obvious continuous projection from $L_0 \setminus K_0^\circ$ to its boundary so we notice that it moves every point by at most $e_0$ and compose it with $\phi_0$ to get $\psi_0$.

Now the boundary of $L_0$ may intersect infinitely many disks (of radius less than $d_0/6$), but only a finite number of disks of radius larger than $e_0$.

Let $K_1$ be the reunion of $L_0$ and the disks of radius larger than $e_0$ it intersects. Since there were finitely many disks, $K_1$ is compact and stays within $3d_0/6$ of $K_0$. For every disk added we project the region that sticks out to its boundary, for example by choosing a point of its circumference inside $L_0$ and projecting radially away from that point. Doing so moves every point by at most $2d_0/6$ and we compose it with $\psi_0$ to get $\phi_1$.
The boundary of $K_1$ is made of finitely many arc circles, with no two consecutive arc radius larger than $e_0 \le d_0/6$.

Now we pick $d_1 \le \min (d_0/6, d(K_1, e_0))$ such that the $d_1$-thickening from an arc of $K_1$ doesn't meet the thickening from any other arc, except for consecutive arcs. Then $B(K_1,d_1) \subset B(K_0,d_0)$

Now the goal is to build a simply connected compact set $K_2$ and a distance $d_2$ such that $K_1 \subset K_2^\circ \subset K_2 \subset B(K_2,d_2) \subset B(K_1,d_1)$ ; $K_2$'s boundary is made of finitely many arc circles, where among two consecutive arc circles, one of them has radius $\le d_1/6$. Together with a projection from $K_2 \setminus K_1$ to the boundary of $K_2$ that moves points by at most some $O(e_0)$.

Then we choose $e_1 \le \min(d_1/6,d(K_1,d_1/6))$ such that $L_1 = B(K_1,e_1)$ is not tangent to any disk. Moving the boundary of $K_1$ to the boundary of $L_1$ can also be done radially from each center of the disks from each arc, except at the corner cases where two bands overlap or go backwards. We can send all those points in a straight line onto the new corner and we still have a continuous projection that moves every point by at most $e_1$.

Here there may be (possibly infinitely many, possibly zero) disks of radius less than $d_1/6$ that touch the $e_1$-thickening of consecutive arcs. So at each corner we push the area in the corner to the outer border of a hypothetical disk of radius $d_1/6$ tangent to both arcs. Such a disk stays within $d_1$ of $K_1$ (because $e_1+2d_1/6 < d_1$) and so it can't touch non consecutive arcs. We move the points for example by projecting in a direction parallel to the tangent line in the corner. The maximum displacement is bounded by the circumferences of the two smaller two disks out of three, so it's a $O(d_1/6+(e_0+e_1)) = O(e_0)$. Let $M_1$ be the resulting set after pushing out every corner.

Now the boundary of $M_1$ may intersect infinitely many disks (of radius less than $d_1/6$), but only a finite number of disks of radius larger than $e_1$. Among two consecutive arcs of $M_1$, one of them has radius less than $d_1/6$. Let $K_2$ be the union of $M_1$ and the disks of radius larger than $e_1$ it intersects, again along with the possible corners filled up. Filling the corner displaces points by a $O(e_1+d_1/6)$, and then pushing up to the outer border of the new disks by $2e_1$.

$K_2$ is simply connected, compact and stays within $5d_1/6$ of $K_1$ so it is still within $d_1$ of $K_1$. Its border is made of finitely many arcs, with no two consecutive arc radius larger than $d_1/6$

And so on.

At every step, we get $K_i \subset K_{i+1}^\circ \subset K_{i+1} \subset B(K_{i+1},d_{i+1}) \subset B(K_i,d_i)$, and every disk of radius greater than $e_i$ is either completely outside $B(K_{i+1},d_{i+1})$ or completely inside $K_{i+1}$. We take $T_n = \cup K_i$

From $\phi_i$ to $\phi_{i+1}$ we only displace points by at most some $O(d_i)$. Since this sequence converges to $0$ geometrically, the sequence $(\phi_i)$ converges uniformly to a continuous projection $f_n$ Since the image of $f_n$ stays squeezed between $K_{i+1}$ and $B(K_{i+1},d_{i+1})$, $f_n$ pushes out everypoint of $T_n$ (which is not necessarily closed) on its boundary, which doesn't intersect any disk at all.

After covering every disk of $S$ by a thickening $T$, we compose every projection together to get a continuous map $\Bbb R^2 \setminus C \to \Bbb R^2 \setminus S$ where $C$ is a countable set consisting of the center of the disks we selected previously.

Then we apply this map on any path from $A$ to $B$ that doesn't pass through any point in $C$, and we obtain a path from $A$ to $B$ that stays outisde $S$.