This question is based on R. Osserman's proof of the four-vertex-theorem (see here: https://arxiv.org/pdf/math/0609268.pdf ).
Without explaining anything, neither in the original work of Mr. Ossermann, nor in the above cited paper is there a proof for the claim that there exists a minimal circle around a compact set. Of course, assuming that it exists it's easy to see why it is unique, but the existence itself is, according to my intuition, hard.
My own tries:
Consider the compact set $E$ and the set $\mathfrak{D}$ of all closed discs, such that $\forall D \in \mathfrak{D} : D \supseteq E$.
Let $(D_i)_{i \in I}$ be a decreasing family of closed discs in $\mathfrak{D}$. It's enough to show that $D:= \cap _{i\in I} D_i \in \mathfrak{D}$, because then it's shown that each monotonically decreasing sequence of discs has its minimum, therefore with Zorn's Lemma, there would exist a global minimum.
It follows immediately that $D$ is compact and $D \supseteq E$, as $\forall i \in I : D_i \supseteq E $. To show is, that $D$ is a disc.
Assuming $D$ wouldn't be a disc, then for any disc $D$', the set $\tilde{D} := D \Delta D' $ with the symmetric difference operator Delta wouldn't be empty.
Form here I have no idea how to get the contradiction.
Note: I didn't get any hints for how to do it, I'm just reading Ossermann's paper.
If there is a simpler or better proof, I'd like to get the source, as it's not obvious to me like Ossermann states it.
HINT:
Let $r = \inf$ of radiuses of discs containing $A$. Consider a sequence of discs $D(c_n, r_n)$ containing $A$, with $r_n \to r$. Now, $c_n$ is a bounded sequence (why?), so it has a convergent subsequence. We may assume $c_n$ convergent to $c$. Recall that $r_n$ convergent to $r$, and $D(c_n, r_n)$ contains $A$. Now show that $D(c, r)$ contains $A$.