In Baire category theorem it says a locally compact Hausdorff space $S$ is second category.
In the proof, it choose $V_1,V_2\cdots$ are dense open subset of $S$. $B_0$ is an arbitrary nonempty open set in $S$. Then we choose $B_n\not = \emptyset$,$\bar B_n \subset V_n \cap B_{n-1}$. We can choose $\bar B_n$ is compact. Put $K=\cap_{n=1}^\infty \bar B_n$. Because in compact space, every family of closed subsets having the finite intersection property has non-empty intersection. So $K$ is nonempty. And $K\subset B_0,K\subset V_n$. Hence $B_0$ intersects $\cap V_n$
So where the Hausdorffness is used. It seems that it should only be regular space. So where it use that it should be regular?
Best Answer
You say: "We can find open $B_n$ with $\overline{B_n}\subseteq V_n\cap B_{n-1}$ such that $\overline{B_n}$ is compact". But why? Strong local compactness tells me that the open neighbourhood $V_n\cap B_{n-1}$ contains a compact neighbourhood $K$. That means, there is an open $B_n\subseteq K$. I would now love to conclude that $\overline{B_n}$ is also compact - but I can't! Not without further assumptions. If I assume the space is Hausdorff, then $K$ is closed, so $\overline{B_n}\subseteq\overline{K}=K$ and I get to conclude $\overline{B_n}$ is compact. It's worse than that, though: I need to also conclude $\overline{B_n}\subseteq V_n\cap B_{n-1}$. This is again easy if $K$ is closed, but it doesn't follow if $K$ is not closed (which is possible if the space is not Hausdorff).
If I'm not allowed to choose the $B_\bullet$ to be precompact, then the proof breaks down at the end. Specifically, I need to be able to conclude at least one $B_n$ has $\overline{B_n}$ compact.
Consider the example of John Doe. $X=\Bbb N$ with the cofinite topology and we define $V_n:=\Bbb N\setminus\{n\}$ for every $n\in\Bbb N$. The closed sets are precisely the finite sets; an infinite set's closure is always $\Bbb N$ itself. So if I want $\overline{B_n}\subseteq V_n\cap B_{n-1}$ I must at least have every $B_n$ finite (then compactness would also follow nicely). But for $B_n$ to be open, it must be infinite. So I can actually never have a $B_n$ whose closure is (compact and) contained in $V_n$.