TITLE: Prob. 2 (d), Sec. 27, in Munkres' TOPOLOGY, 2nd ed: If $A$ is compact and $U$ is an open set containing $A$, then . . .
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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...