Does a second-countable, locally compact, Hausdorff space admit a countable basis of pre-compact open sets? Let $X$ be a such space. I know that for a Hausdorff space, being locally compact is equivalent to having a basis of pre-compact open sets. But how do I prove that $X$ can be covered by countably many such sets, by using second-countability of X? Thank you.

Every locally compact Hausdorff space is (completely) regular. Every second-countable regular Hausdorff space is metrisable, and hence separable. Any locally compact metrisable space admits a compatible metric, each of whose (proper) open balls is precompact.
Thus if $X$ is locally compact, Hausdorff and second-countable, then it is Polish. It admits a compatible, but possibly noncomplete, metric $d$ for which there is a countable subset $\{x_n\}_{n\in\mathbb{N}}$ such that $\{B_{\frac{1}{m}}(x_n)\mid m,n\in\mathbb{N}\}$ is a countable base of precompact sets for $X$.
