Every locally compact, Hausdorff, second-countable space has a countable basis consisting of open sets with compact closure 
I'm trying to understand the proof of this lemma. My problem lies in the existence of $V_{n(x)}$. As far as I know (from Munkres' topology), there should be a neighborhood $U_x$ of $x$ such that $\overline{U_x}$ is compact and $\overline{U_p}\subseteq V_{n(x)}$. Can anyone show me how? Thanks. 
 A: That there is an open set $U_x$ containing $x$ with $\overline {U_x}$ compact is a theorem on locally compact Hausdorff spaces. There there exists some $V_n=V_{n(x)}$ contained in $U_x$ and containing $x$ because $(V_n)$ is a base. Since $V_{n(x)} \subset U_x$ it follows that the closure of $V_{n(x)} $ is compact. Since $(V_{n(x)})$ is a sub-collection of $(V_n)$ is is countable. That finishes the proof.  
Edit: as pointed out by Steve it is not clear that $(V_{n(x)})$ is a base. So the proof has to me modified as follows: let $U$ be any open set and $x \in U$. Pick $U_x$ as above. Then there exists $V_{n(x)}$ such that $x \in V_{n(x)} \subset U \cap U_x$. Rest is now clear. 
A: $\{V_n\}$ is a basis for the topology. Then by definition, given any open set $U$ and a point $x\in U$, we can find a basis element $V_{n(x)}$ such that $x\in V_{n(x)}\subset U$.
A: 
My problem lies in the existence of $V_{n(x)}$.

Given any open set $U$ at all around $x$, the phrase "$\{V_n\}$ is a basis of the topology" means that one of the $V_n$'s can get inside of $U$ and around $x$ too. This explains the existence of $V_{n(x)}$.
