The task:
S is a non-empty closed subset of $\mathbb{R}^n$ equipped with the Euclidean metric.
Take $a \not \in S$. Show $\min \{ d(x,a) \ | \ x \in S \}$ exists.
Below is my attempt. I wanted to ask for feedback on it. To me it appears airtight, but perhaps I glossed over some logical hole.
Proof:
As $a \not \in S$, there exists a real $\delta > 0$ such that $B(p, \delta) \subset E \smallsetminus S $.
Therefore for any $x \in S$ we have $d(x,a) \geq \delta$.
Thus the set $ \{ d(x,a) \ | \ x\in S \} $ is non-empty and bounded below. Completeness of reals provides the existence of $\alpha = \inf \{ d(x,a) \ | \ x\in S \}\geq \delta > 0 $.
Construct a sequence $(x_n)_{n=1}^\infty$ of points in $S$, such that for each natural $n$ we have
$\alpha \leq d(x_n, a) < \alpha + \frac 1n$ .
Note this means the sequence $\big(d(x_n,a)\big)_{n=1}^\infty$ converges to $\alpha$.
Moreover this means each point $x_n$ lies in the closed ball $\overline{B}(a, \alpha + 1)$.
In fact, all the points in $(x_n)_{n=1}^\infty$ lie in the closed and bounded set $S \cap \overline{B}(a, \alpha + 1)$, which is compact. Therefore there is a convergent subsequence $x_{n_k} \to p \in S $.
The function $f: S \to \mathbb{R}$ which sends $x \mapsto d(x,a) $ is continuous, so it maps convergent sequences to convergent ones.
Hence $f(x_{n_k}) = d(x_{n_k}, a)$ converges to $f(p) $. But we saw above it also converges to $\alpha$. Therefore $\alpha = f(p)$.
This shows $\alpha = \inf \{ d(x,a) \ | \ x\in S \} = \min \{ d(x,a) \ | \ x\in S \}$. We are done. $\blacksquare$
Best Answer
That's fine. You could compactify (no pun intended) the very argument you made (and at the same time drop the condition $a\notin S$):
As $S$ is non-empty, pick $s\in S$. Then $S':=S\cap \overline{B(a,d(a,s)+1)}$ is non-empty, closed, and bounded, hence compact. Hence for the continuous function $x\mapsto d(a,x)$, there exists some $p\in S'$ that minimizes it. From $d(a,x)\ge d(a,s)+1\ge d(a,p)$ for $x\in S\setminus S'$, we see that $p$ is at the same time minimizing $d(a,x)$ on all of $S$.