Let $(X,\tau)$ be a second countable, locally compact Hausdorff space.
Theorem:
If $S \in \tau$, then $S$ is $\sigma$-compact.
How do I show this statement? The following is what I have tried:
Let $\mathcal{E}$ be a countable base for $(X,\tau)$. For each $x \in S$ there exists a compact neighbourhood $V(x), x \in V(x) \subseteq S$.(This is a theorem that I'm allowed to use) Then $S = \bigcup_{x \in S} V(x)$, but how do I show that there exists a finite cover?
You can restrict to compact neighborhoods of the form $\overline V$ where $V\in\cal E$. This follows from the space being locally compact and Hausdorff. This way you would get $V=\bigcup_{x\in S}\overline{V(x)}$ and this union is countable as there are only countably many elements in $\cal E$.