Exercise 2(d), Section 27 of Munkres’ Topology 
Let $X$ be a metric space with metric $d$; let $A \subset X$ be nonempty.
(d) Assume that $A$ is compact; let $U$ be an open set containing $A$. Show that some $\epsilon$-neighborhood of $A$ is contained in $U$.

My attempt:
Approach(1): Since $U\in \mathcal{T}_d$ and $A\subseteq U$, we have $\forall a\in A, \exists r_a \gt 0$ such that $B_d(a, r_a) \subseteq U$. $B=\{ B_d(a, r_a)| a\in A\}$ is an open cover of $A$. $A$ is compact. By lemma 27.5, $\exists \delta \gt 0$ such that $E\subseteq X$ and diam $(E) \lt \delta$ $\Rightarrow$ $E\subseteq B_d(x, r_x)$, for some $x\in A$. Take $\epsilon =\frac{\delta}{4}$. By exercise 2(c) section 27, $U(A,\epsilon)=\bigcup_{a\in A} B_d(a, \epsilon)$. Let $x,y\in B_d(a, \epsilon)$; $a\in A$. Then $d(x,y)\leq d(x,a)+d(a,y)\lt 2\epsilon =\delta /2$. So diam$(B_d(a, \epsilon))\leq \delta /2 \lt \delta$. Thus $B_d(a, \epsilon)\subseteq B_d(z, r_z)\subseteq \bigcup_{a\in A} B_d(a, r_a)$, for some $z\in A$. Hence $ U(A,\epsilon)=\bigcup_{a\in A} B_d(a, \epsilon)\subseteq \bigcup_{a\in A} B_d(a, r_a)\subseteq U$. Our desired result. Is this proof correct? Did I precisely use lemma 27.5?
Approach(2): Prob. 2 (d), Sec. 27, in Munkres' TOPOLOGY, 2nd ed: If $A$ is compact and $U$ is an open set containing $A$, then . . ..
Approach(3): https://math.stackexchange.com/a/2939391/861687.
 A: I am guessing that the Lemma 27.5 must be the Lebesgue number lemma. Here is the Lebesgue number lemma (wording might be slightly different from Munkres's book. I don't have a copy with me right now.)
Lebesgue Number Lemma Let $(Y, d)$ be a compact metric space and let $\mathcal{U}$ be an open cover of $(Y, d)$. Then, there exists $\delta>0$ such that for every $E\subseteq X$ such that $diam(E)<\delta$, there exists a set $U\in \mathcal{U}$ such that $E\subseteq U.$
If you want to use it exactly like this, you will need to modify your argument a little. The metric space $(X, d)$ is not necessarily compact. But we have a compact set $A\subseteq X$. It can obviously be thought of as a compact metric space in its own right. But the covering $\{B_d(a, r_a): a\in A\}$ is not a cover of $A$ in $(A, d)$  (Why?). To remedy the situation, you can consider a modified cover, say, $\mathcal{C}:=\{B_d(a_i, r_{a_i})\cap A: a_i\}$ where $\{a_i\}$ is a finite collection of elements from $A$  such that $A\subseteq \cup_i B_d(a_i, r_{a_i}/4)$.
Now you can apply Lebesgue Number Lemma to this covering. This yields a $\delta>0$ such that every $E\subseteq A$ with diameter less than $\delta$ will be contained in $B_d(a_i, r_{a_i}/4)\cap A\subseteq B_d(a_i, r_{a_i}/4)$ for some $i$. Take $\epsilon<\min\{\delta, r_{a_i}\}/4$ (may not be necessary) and claim that $\epsilon$ works for you. That is, take any $x\in A$ and consider the set $E=B_d(x, \epsilon)$. You need to show that $E\subseteq \cup B_d(a_i, r_{a_i}).$
To this end, note that $E\cap A$ has diameter less than $\delta$ and hence by Lebesgue number lemma we know that $E\cap A\subseteq B_d(a, r_a/4)\cap A$ for some $a\in A.$ Now take any $z\in E$ and note that $d(a, z)\leq d(z, y)+d(y, a)$ for some $y\in E\cap A$. It follows that $d(a, z)\leq \epsilon+r_{a}/4\leq r_a/2.$ In particular, $E\subseteq B(a, r_a).$
