Let $(X,d)$ be a complete metric space. Suppose $B \subset X$ is compact.
Prove that for every $a\in X$ the minimum $\min_{b\in B} d(a,b)$ exists.
I'm pretty sure you can do this by just using the fact that the metric is a continuous function on $X$ and so $f(x)=d(x,a)$ is continuous, and so $f(B)$ is compact subset of $\mathbb{R}$ and thus attains a minimum, $z$. We can take any element of $f^{-1}(z)\neq \emptyset$. However, I wanted to try it another way and was hoping I could have anyone validate it, catch my mistakes, or show me how these are more similar then I realize, etc.
My alternative attempt:
Suppose for some $a\in A$ this minimum did not exist. Consider the famly of open sets $\{U_{b'}\}= \{x\in X-b' \mid d(a,x) > d(a,b')\}$ , where $b'$ ranges over all of $B$. These sets are open as they are the intersection of two open sets (by definition of the topology on $X$). If they do not cover $B$, then $\exists$ some $b''\in B$ such that $b'' \notin \bigcup_{b'\in B} U_{b'}$, which implies $d(a,b'')\leq d(a,b')$ for all $b'\in B$. This contradicts our assumption that no minimum exists.
Thus this collection of sets $\{U_{b'}\}$ does cover $B$, and so by compactness we have a finite subcover: $\{U_{b_1}, \ldots, U_{b_k}\}$. By well-ordering we have a minimum of this finite set of non-negative integers $\{d(a,b_1), \ldots, d(a,b_k)\}$. WLOG let this be $d(a,b_1)$. However, this implies that $b_1 \notin U_{b'}$ for all $b' \in B$, which is another contradiction. So we must have a minimum for every $a'\in A$.
You can generalize your "alternative attempt" to an arbitrary collection of real-valued continuous functions $f_{\alpha}:X\rightarrow\mathbb{R}$. The set $\left\{x : f_{\alpha}(x)>f_{\alpha}(b')\right\}$ is the preimage of the open set $(f_{\alpha}(b),+\infty)$, hence open by the continuity of $f_{\alpha}$. If there exists $b''\in B\setminus\bigcup_{b'}U_{b'}$, then $f_{\alpha}(b'')\leq f_{\alpha}(b')$ for every $b'\in B$, which contradicts our hypothesis that $f_{\alpha}$ does not attain its minimum on $B$.