The Baire category theorem is theorem 48.2 in Munkres. The proof of the theorem, and so also a proof adequate for this problem, is found there (page 296 in my edition). I used the proof outlined there to construct my proof for problem 27.5.
There is a line of the proof which states
Regularity of $X,$ along with the fact that $A_1$ is closed, enables us to choose a neighborhood $U_1$ of $y$ such that $$\bar{U}_1 \cap A_n = \emptyset,$$ $$\bar{U}_1 \subset U_0.$$
The ideas used in proving theorem 27.7 are needed to justify this step, which justification Munkres leaves out in proving 48.2. Step one of the proof of 27.7 says
We show first that given any nonempty open set $U$ of $X$ and any point $x$ of $X,$ there exists a nonempty open set $V$ contained in $U$ such that $x \notin \bar{V}.$
For problem 27.5, you need to show that for all $n$ there exists a nonempty open set $U_n$ contained in $U_{n-1}$ such that $A_n \cap \bar{U}_n = \emptyset.$ This is the similarity with theorem 27.7.
Bellow is my proof, which I believe to be correct. Perhaps it will clarify what I mean.
Let $X \neq \emptyset$ be a compact Hausdorff space and let $\{A_n\}_{n \in \mathbb{N} }$ be a countable collection of closed subsets of $X.$ Assume that $\text{Int} A_i = \emptyset \,\forall\, i.$ Let $U_0$ be any nonempty open set. Then $U_0 \not\subset A_1,$ so we can choose some $y_1 \in U_0 \setminus A_1.$ Since $X$ is Hausdorff and $A_1$ is compact, by lemma 26.4 we can choose neighborhoods $U_1$ of $y_1$ and $V_1$ containing $A_1$ such that $U_1 \subset U_0$ and $U_1 \cap A_1 = U_1 \cap V_1 = \emptyset \implies \bar{U}_1 \cap A_1 = \emptyset.$ Since $U_{n-1} \not\subset A_n,$ we can follow a similar procedure for $n>1$ to choose $y_n \in U_{n-1} \setminus A_n$ and then choose an open set $U_n$ around $y_n$ such that $\bar{U}_n \cap A_n = \emptyset$ and $\bar{U}_n \subset U_{n-1}.$ Since $X$ is compact, by theorem 26.9 $\exists\, z \in \bigcap_{n \in \mathbb{N} } \bar{U}_n.$ This $z$ is not contained in any of the $ A_n,$ so $z \notin \bigcup_{n \in \mathbb{N} }A_n.$
This shows that every open set must contain an element not contained in $\bigcup_{n \in \text{N} } A_n.$ It follows that $\bigcup_{n \in \mathbb{N} } A_n$ can contain no open sets.