Proving that a compact Hausdorff space is a Baire space. This is exercise 27.5 in Munkres:

Let $X$ be a compact Hausdorff space; Let $\{A_n\}$ be a countable collection of closed sets of $X$. Show that if each set $A_n$ has empty interior in $X$, then the union $\bigcup A_n$ has empty interior in $X$. [Hint: Imitate the proof of Theorem 27.7.]

Here is Theorem 27.7:

Let $X$ be a nonempty compact Hausdorff space. If $X$ has no isolated points, then $X$ is uncountable.

The proof of Theorem 27.7 boils down to this:


*

*Show that given any nonempty open set $U$ of $X$ and any $x\in X$, there exists a nonempty open set $V$ contained in $U$ such that $x\notin\overline V$.

*Given $f:\mathbb Z_+\to X$, use 1. to construct a sequence of points $x_n$ all distinct from $x$. It follows that $f$ is not surjective, and hence $X$ is uncountable.


I don't see any relation between the exercise and the proof of the theorem. The theorem constructs a sequence of points, while the exercise requires me to show that a set has empty interior. Any hints on how to relate these two would be greatly appreciated (I'd like to complete the proof myself).
 A: Suppose that $O \subseteq \bigcup_n A_n$ is non-empty open.
$A_1$ has empty interior and is closed, so there is some $O_1$ non-empty open with $\overline{O_1} \subseteq O$ such that $O_1 \cap A_1 = \emptyset$.
$A_1 \cup A_2$ also has empty interior and is still closed, so there is some non-empty open $O_2$ such that $O_2 \subseteq \overline{O_2} \subseteq O_1$ and $O_2 \cap (A_1 \cup A_2)=\emptyset$.
Continue finding $O_n$ this way and note that $\bigcap_n \overline{O_n}$ is non-empty by compactness and see what contradiction you find. 
With a minor adaptation at the start, this also works for locally compact Hausdorff spaces.
The common theme with the older proof is the decreasing open sets intersecting to achieve a "countable goal" when finite goals are achievable.
A: 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.

