Chebyshev sets in finite dimension are closed and convex

Prove a finite-dimensional converse to the “best approximation theorem”: Let $K$ be a subset of a finite-dimensional Hilbert space $H$ which satisfies the following property: for each $x \in H$ there exists a unique point $y \in K$ such that $d(x, y) = d(x, K)$ (sets with this property are known as Chebyshev sets). Then the set $K$ is closed and convex.

It would useful to point why this result is for finite dimensional spaces only.

Here's my proof: it's pretty elementary but in the proof of convexity there are a lot of named points, so a drawing may help (included at the end).

Proof of closedness: Assume $K$ is Chebyshev, and choose $\alpha\in K^c$. There is then $\beta\in K$ such that $d(\alpha,\beta)=d(\alpha,K)$. In particular $\alpha \not = \beta$ and so $d(\alpha, K)>0$. Thus $\alpha$ is an interior point of $K^c$. Since $\alpha$ was arbitrary, $K^c$ is open and thus $K$ is closed.

Proof of convexity: Assume now for a contradiction that $K$ is not convex. Then there are $x,y\in K$ such that the line segment connecting them contains points not in $K$. Since $K$ is closed, without loss of generality we may assume that there are no points of $K$ between $x$ and $y$ (the intersection of $K^c$ and the line segment is a nonempty countable collection of open intervals, so we may choose one such interval).

Let $z:=(x+y)/2$. By choice of $x,y$, we have $z\in K^c$. Since $K$ is Chebyshev there must be a point $w\in K$ with $0<d(z,K) = d(z,w)<d(z,x)$

Consider the line $L:=\{\lambda w + (1-\lambda)z:\lambda \in \mathbb{R}\}$. We now invoke the converse of the theorem we are trying to prove, to conclude that $L$ is a Chebyshev set. So there are unique projections $x' := \lambda_x w + (1-\lambda_x)z$ and $y' := \lambda_y w + (1-\lambda_y)z$ of $x$ and $y$ onto $L$ respectively.

I claim that $\lambda_x=-\lambda_y$. This follows from the "angle-angle-side" congruence theorem of elementary geometry, i.e., $\triangle xzx' \cong\triangle yzy'$. In particular then at least one of $\lambda_x,\lambda_y$ is $\leq 0$ (we can also get to this fact by contradiction using the triangle inequality). By symmetry, say $\lambda_x\leq 0$. That is, $x'$ is not between $z$ and $w$, nor is it on the $w$ side of the line.

From what has preceded, $0<d(x,x')\leq d(x,z)$. By the Pythagorean theorem we may now find a point on $L$ (and in $K^c$) equidistant to $w$ and $x$, and with $\lambda<0$. This by itself is not a contradiction, but what it means is if we split $\{\lambda w + (1-\lambda)z:\lambda <0\}$ into sets $A$ and $B$, where points of $A$ project onto $K$ at $w$ and points of $B$ project onto some other point of $K$, then $B\not = \emptyset$.

Finally let $s=\lambda_sw + (1-\lambda_s)z$, where $\lambda_s=\inf\{\lambda:\lambda<0,\lambda w+(1-\lambda)z \in B\}$. In other words, $s$ is the closest point to $z$ on $L$ (in the direction away from $w$) which either is in $B$ or the limit of a sequence in $B$. Indeed I will show that $s\in B$ presently. Let $\{s_k\}_{k\geq 1}$ be a sequence of points in $B$ converging to $s$. There is then a corresponding sequence $\{t_k\}_{k\geq 1}$ of projections onto $K$, that is, $s_k \mapsto t_k$ is the projection map.

All of $\{t_k\}_{k\geq1}$ is contained within a suitably large ball around $s$. Recall $K$ is closed, so by the Bolzano Weierstrass property, there is a convergent subsequence $\{t_{k_i}\}_{i\geq 1}$, where $t_{k_i}\to t$ as $i\to \infty$. Indeed, here is where the proof breaks in the infinite dimensional case, since such a closed, bounded subset is not totally bounded. By the closure of $K$, $t\in K$, and by continuity of the distance function, we also have $s$ projects onto $t$. So $s\in B$.

From here the contradiction is apparent, since it must be the case that $d(s,t) = d(s,w)$. We have established that $s$ projects onto $t$, but $t$ is not the unique closest point of $K$ to $s$, contradicting the fact that $K$ is Chebyshev.