This subcollection of the base is also a base Let $B_n$ be a countable base of a locally compact Hausdorff space $X$. I am trying to show that $S = \{B_n \mid \overline{B_n} \text{ is compact} \}$ is also a base.
I imagine the proof to be short and simple but I can't seem to find it. 
The goal: Given an open set $U$ in $X$ I want to show that there are $B_n \in S$ such that $U = \bigcup_n B_n$. I'm not sure how to use that $X$ is Hausdorff. 
My thoughts on this: I know that a compact subset of a Hausdorff space is closed but since $\overline{B_n} $ is already closed this doesn't come in handy. Similarly, if $u$ is any point in $U$ the local compactness gives a compact neighbourhood $N$ of $u$. But I don't see how to relate $N$ to the elements in $S$.
Hence my question: How to prove this?
 A: The important point is that in a locally compact Hausdorff space, every point has a neighbourhood basis consisting of compact neighbourhoods.
Given that, it is clear that for every $x\in U$ there is a compact neighbourhood $x \in \overset{\Large\circ}{K} \subset K \subset U$, and since $\{B_n\}$ is a basis, there is an $n_x$ with $x \in B_{n_x} \subset K$. Then $\overline{B_{n_x}}$ is clearly compact, hence $B_{n_x}\in S$, and the criterion for $S$ to be a basis is verified.
So let's see that the compact neighbourhoods of a point $x$ form a neighbourhood basis. Let $K$ a compact neighbourhood of $x$, and $U$ any open neighbourhood of $x$. We want to see that $U$ contains a compact neighbourhood $L$ of $x$. Let $V = U\cap \overset{\Large\circ}{K}$. Then $V$ is an open neighbourhood of $x$ contained in $U$, and $K\setminus V$ is a compact set disjoint from $x$. By the Hausdorff property, we find disjoint open sets $W$ and $Z$ such that $x \in W \subset V$ and $K\setminus V \subset Z$. Then $L := \overline{W} \subset \overline{V} \subset K$, so $L$ is a compact neighbourhood of $x$, and $L \cap Z = \varnothing$, so in particular $L\cap (K\setminus V) = \varnothing$, which means $L \subset V\cup (X\setminus K)$, but since $L\subset K$, we have $L \subset V \subset U$, as desired.
