Compact Hausdorff space is of second category

Question

Let $X$ be a compact Hausdorff space, prove $X$ is of second category.

I found a proof of this theorem in the case of locally compact Hausdorff spaces.

Let $E_n$ be open and dense in $X$, locally compact and Hausdorff. Let $$D= \bigcap_{i=1}^{\infty} E_i$$ We must show that $D$ is dense, as this will imply that $X$ is of second category.

Let $U$ be a non-empty open subset of $X$. For $D$ to dense we must have that $U$ intersects $D$. Now $U \cap E_1$ is open, and thus there exists open set $V_1$ such that $\operatorname{cl}(V_1) \subset U \cap E_1$, by Hausdorff and local compactness.

I know that there exists an open set $V_1 \subset U \cap E_1$, since intersection of $U$ and $E_1$ is open and $E_1$ is dense. However, Why does local compactness and Hausdorff imply that closure of such a set will be a subset?

Answer 1

$X$ is a locally compact Hausdorff space, so every point has a local base of open nbhds with compact closures. Suppose that $x\in X$ and $U$ is an open nbhd of $x$. Then there is an open $B$ such that $x\in B\subseteq U$, and $\operatorname{cl}B$ is compact. But then $\operatorname{cl}B$ is a compact Hausdorff space and therefore regular, so there is an open nbhd $V$ of $x$ such that $\operatorname{cl}V\subseteq B\subseteq U$.

Answer 2

But any compact space $ X $ is locally compact, since $ X $ is a compact neighborhood of $ x $ for all $ x\in X $. So the proof will still apply.

Edit: Any compact Hausdorff space is normal. See also http://topospaces.subwiki.org/wiki/Compact_Hausdorff_implies_normal. Let $ x\in U\cap E_1$. Now note that $\{x\} $ and $ X-U\cap E_1 $ are disjoint closeds. Hence we find $ V_1, V_2 $ disjoint opens containing $\{x\} $ and $ X-U\cap E_1 $ respectively. Now $\bar {V_1} \subset U\cap E_1 $.

I'm not quite sure about how to attack the only locally compact case. We could try to find a big enough compact neighborhood such that it contains $U\cap E_1$. Then we have a compact Hausdorff subspace in which we can use the above argument. I'm not sure if we can always find this though.