Proof of property of unique nearest point equal to convex.

Question

In the picture below, I really don't know how to use the compactness to get there will be a ball $C$ with maximal radius.

I report this book for my classmate and teacher. I tell they the $r$ has a upper bound because the restricting of $B((x+y)/2,\rho)$ and concave of $A$. But they think the imagination of geometry is not reliable.Then, I explain this question by three points determinate a circle and three points only can be in three close set, but I can proof the determinate is continue (is hard to compute).I really will be crazy!

Picture below is from the 10th page of book in the red picture.Forgive me can't give any hyperlink.

Answer 1

I think, the compactness argument is meant in the following sense:

1. there is a $\bar r > 0$, such that for all balls $B(z, r)$ with $B \subset B(z,r)$, but $x,y \not\in B(z,r)$, we have $r \le \bar r$. This can be justified by a simple, two-dimensional geometric argument in the affine hull of $x,y,z$.
2. Take the smallest $\bar r$, such that all balls $B' \in \mathcal{B}$ have a radius smaller or equal to $\bar r$.
3. There is a sequence of balls $B(z_n, r_n)$ satisfying $B \subset B(z_n,r_n)$ and $A \cap \operatorname{int}(B(z_n, r_n)) = \emptyset$ and $r_n \nearrow \bar r$. By a compactness argument, a subsequence of $z_n$ converges to some $z$.
4. It is not hard to show that $B \subset B(z, \bar r)$ and $A \cap \operatorname{int}(B(z,\bar r)) = \emptyset$.