Proof of existence of the point

Question

Let $C$ be a closed, convex set and $x \notin C$. Then exists one and only one point $c \in C$ such that $\|x - c\| = min_{y \in C} \|x - y\| = d(x, C)$. The proof of uniqueness by reductio ad absurdum is clear. But the proof of existence confuses me. Namely, we take a $v \in C$ arbitrary and observe the set $$V = \{y \in C : \|x - y\| ≤ \|x - v\|\} = C \cap \bar{B}_{x-v}(x).$$I realize the set $V$ is closed and convex (as an intersection of such sets), but why is $V$ compact? Further, the proof observes $\phi : V \rightarrow \mathbb{R} , \phi(y) = \|x - y\|$ which is continuous on $V$, thus it reaches its minimum and the claim is proved. But why is $V$ compact?

Answer 1

A closed subset of a compact set is always compact. So, $C$, being a closed set, intersects with $\overline{B}_{\|x - v\|}(x)$, which is a compact set, producing $V$, a closed subset of the compact set $\overline{B}_{\|x - v\|}(x)$, and hence is compact. So yes, $V$ is indeed compact.

Very similarly, you can also appeal to the Heine-Borel theorem. $V$ is the intersection of closed sets, and hence closed. Since it lies inside a closed ball, it is also bounded.