Let $(T_n)$ be a basis of a locally compact space $E$. For each $x\in E$, there exists a compact neighborhood $C_x$ of $x$. So, there exists an index $n_x$ such that $x\in T_{n_x}\subset C_x$. Thus $T_{n_x}$ is relatively compact for each $x\in E$. I want to prove that $(T_{n_x})_{x\in E}$ forms a basis for the open sets of $E$.
Let $x\in E$ and $V$ be a neighborhood of $x$ in $E$. We have to show that there exists $y\in E$ such that $x\in T_{n_y}\subset V$. By assumption, there exists an index $n$ such that $x\in T_n\subset V$ and also $x\in T_{n_x}$. I don't know where to go from here. Any suggestions?
I’d go the easy route: define
$$\mathcal{B}=\{T_n \mid \overline{T_n} \text{ is compact}\}$$
And try to show this is a base.
To this end let $O$ be open in $E$ and $x \in O$. Let $C_x$ be a compact neighbourhood of $x$ and let $T_n$ be any of the base $(T_n)$ we started with such that $x \in T_n \subseteq \operatorname{int}(C_x) \cap O$. As in particular $T_n \subseteq C_x$ we have that $\overline{T_n}$ is indeed compact and so $T_n \in \mathcal{B}$ and also $x \in T_n \subseteq O$ as required. This shows that $\mathcal{B}$ is indeed a base for $E$.