a subset of $\mathbb R^2$ for which minimum distance is reached exactly once is convex and closed.

Question

Let $A\subseteq \mathbb R^2$ be a set such that for every $x\in \mathbb R^2$ there is a $y\in A$ with $d(x,y)

Proposed solution:

Clearly the set must be closed. Let $d(x,A)$ be the distance from a point $x$ to $A$, let $R$ be a closed rectangle not contained in $A$ and let $M$ be the maximum of $d(x,A)$ for points in the rectangle, if $d(x,A)=M$ for $x\in R$ we must have that $x$ is a corner. Proof: suppose not, consider the point $z$ that is farthest away from $x$ and move away slightly from $z$ in a straight line , you'll find a point $x'$ with $d(x',A)>d(x,A)$ (you need to use that $A$ is closed but it works out). Suppose $A$ is not convex, pick points $a$ and $b$ in $A$ such that some $c$ in between is not in $A$. If you consider a rectangle in which one side is $ab$ and the other has las length less than $d(c,A)$ you get a contradiction.

I'm looking for verification and alternative proofs.