Here is Prob. 2, Sec. 27, in the book Topology by James R. Munkres, 2nd edition:
Let $X$ be a metric space with metric $d$; let $A \subset X$ be nonempty.
(a) Show that $d(x, A) = 0$ if and only if $x \in \overline{A}$.
(b) Show that if $A$ is compact, $d(x, A) = d(x, a)$ for some $a \in A$.
(c) Define the $\epsilon$-neighborhood of $A$ in $X$ to be the set
$$ U(A, \epsilon) = \{ \ x \in X \ \vert \ d(x, A) < \epsilon \ \}. $$
Show that $U(A, \epsilon)$ equals the union of the open balls $B_d(a, \epsilon)$ for $a \in A$.(d) Assume that $A$ is compact; let $U$ be an open set containing $A$. Show that some $\epsilon$-neighborhood of $A$ is contained in $U$.
(e) Show the result in (d) need not hold if $A$ is closed but not compact.
This and this are two Math SE posts on this problem. And, here is also a solution to this problem.
I think I'm clear on parts (a) through (c) of this problem. So here I'll give my attempt at part (d).
My Attempt:
First, some notation:
For any point $x \in X$, we define
$$ d(x, A) \colon= \inf \{ \ d(x, a) \ \vert \ a \in A \ \}. \tag{Definition A} $$
And, for any point $p \in X$ and for any real number $\delta > 0$, we define
$$ B_d (p, \delta) \colon= \{ \ x \in X \ \vert \ d(x, p) < \delta \ \}. \tag{Definition B} $$As $U$ is an open set in $X$ with the metric topology determined by the metric $d$, so, for every element $u \in U$, there exists a real number $\epsilon_u > 0$ such that
$$ B_d \left( u, \epsilon_u \right) \subset U. $$
[Refer to Sec. 20 in Munkres, especially the portion of the section preceding Example 1.]In particular, as $A \subset U$, so, for every element $a \in A$, we can find a real number $\epsilon_a > 0$ such that
$$ B_d \left( a, \epsilon_a \right) \subset U. \tag{1} $$
For each such $\epsilon_a > 0$, let us choose a real number $\delta_a$ such that
$$ 0 < \delta_a \leq \frac{\epsilon_a}{2}. \tag{2} $$Now let us consider the collection
$$ \left\{ \ B_d \left( a, \delta_a \right) \ \vert \ a \in A \ \right\}. $$
This is a collection of open sets of $X$ whose union contains the set $A$; that is, this collection is a covering of $A$ by sets open in $X$. So, by Lemma 26.1 in Munkres, there is some finite sub-collection of this collection that also covers $A$. That is, there exist points $a_1, \ldots, a_n \in A$ such that
$$ A \subset \bigcup_{j=1}^n B_d \left( a_j, \delta_{a_j} \right). \tag{3} $$Let us now put
$$ \epsilon \colon= \frac{1}{2} \min \left\{ \ \delta_{a_1}, \ldots, \delta_{a_n} \ \right\}. \tag{4} $$
This $\epsilon > 0$ of course, by virtue of (2) above.Now from Part(c) we have
$$ U (A, \epsilon) = \bigcup_{a \in A} B_d(a, \epsilon). $$
Let us pick an arbitrary point $x$ in $U(A, \epsilon)$. Then as
$$ x \in \bigcup_{a \in A} B_d(a, \epsilon), $$
so by the definition of the union of sets there exists a point $a_* \in A$ such that
$$ x \in B_d \left( a_*, \epsilon \right), $$
that is such that
$$ d \left( x, a_* \right) < \epsilon, \tag{5} $$
by virtue of (Definition B) above.Now as $a_* \in A$, so by virtue of (3) above, we can conclude that
$$ a_* \in B_d \left( a_k, \delta_{a_k} \right) $$
and so
$$ d \left( a_*, a_k \right) < \delta_{a_k}, \tag{6} $$
for at least one $k = 1, \ldots, n$.
And for this same $k$, using (2), (4), (5), and (6) above, we obtain
$$
d \left( x, a_k \right) \leq d \left(x, a_* \right) + d \left( a_*, a_k \right) < \epsilon + \delta_{a_k} < \delta_{a_k} + \delta_{a_k} = 2 \delta_{a_k} \leq \epsilon_{a_k}. $$
Thus
$$ x \in B_d \left( a_k, \epsilon_{a_k} \right). $$
So from (1) we conclude that $x \in U$.But by our choice $x$ was an arbitrary element of $U(A, \epsilon)$. Therefore we have
$$ U(A, \epsilon ) \subset U. $$
Is this proof correct? If so, then is each and every step of this proof clear enough too? If not, then where is it lacking?
Best Answer
It's quite detailed and seems correct to me.
Another approach: $f:x \to d(x,X\setminus U)$ is continuous. As $A \subseteq U$ and $X\setminus U$ is closed, we know that $f(x)>0$ for all $x \in A$. By compactness $\min f[A]$ exists. Let $\varepsilon = \min f[A]>0$.
Then a small argumentation will show that this $\varepsilon$ is as required. (I believe your second linked answer also follows this approach). I believe the continuous distance argument is actually better and builds on stuff that's already been shown so is not more complicated. The whole point of having a body of such results is to make for more convenient proofs later, instead of reducing all compactness proofs to long indexing exercises with covers and finite subcovers...