Let $A$ be an open bounded subset of euclidean $n$-space $\mathbb{R}^n$. For $x\in A$, let $r=r(x)$ be the maximal radius such that the ball centred at $x$ with radius $r=r(x)$ is contained in $A$, i.e. $r(x)$ is the maximal $r>0$ satisfying $B_r(x) \subset A$. N.B. the radius $r(x)$ coincides with distance-to-the-boundary $dist(x,\partial A)$. 

For $x\in A$, let $M_x$ be the maximal ball containing $B_{r(x)}(x)$ and contained in $A$. Thus $M_x$ is the maximal ball satisfying $B_{r(x)}(x) \subset M_x \subset A$. 

***My question:
 If we assume $A$ is open and bounded, then does $x\mapsto M_x$ vary continuously with $x$?***

[![enter image description here][1]][1]

N.B. SRM has provided nice class of counterexamples in his answer, showing the answer is no whenever the medial axis is noncompact and has nontrivial accumulation points (i.e. Cauchy sequences in the euclidean distance). 


N.B. For given $x$, I argue that there exists a *unique* maximal ball $M_x$ satisfying the above conditions.

Remark. For my purposes a positive answer would imply that the centre $m$ of $M_x$ varies continuously with $x$. Thus we would obtain a continuous map $x\mapsto m(x)$ from $x$ to the centre $m(x)$ of the max ball $M_x$.

Remark. I remember reading an article/book of Vitali Kapovich which had some similar constructions, but I cannot recall the specifics. My  goal is to establish continuity of $x\mapsto m(x)$ with mild hypotheses on $A$, i.e. without requiring a $C^1$ boundary. 

Remark. There is a variational characterization of the max-centre defined as the argmax of a convex function on a convex domain. Therefore the argmax is unique element on the boundary of the domain, and varies continuously with respect to the datum. 


  [1]: https://i.sstatic.net/ru3U9.jpg