Any locally compact space is rim compact

Question

Definition

A topological space $X$ is rim compact if for any $x\in X$ there exist a local basis $\mathcal{B}(x)$ such that for any $B\in\mathcal{B}(x)$ the subspace $\partial B$ is compact.

Statement

Any locally compact space is rim compact.

Unfortunately, I can't prove the statement. Anyway, it seems to me that it could be false: however, if we add some hypothesis (e.g hausdorff or tychonoff separability etc...) could it true?

So, could someone help me, please?

Answer 1

If compact sets are closed (so in a Hausdorff space, e.g.) $\partial B$ is a closed subset of $\overline{B}$ so also compact. So in a Hausdorff space having one compact neighbourhood implies we have a local base of compact neighbourhoods (in the general sense), and so then the space is peripherally compact.

If $X$ is $\Bbb R$ in the included point topology wrt $0$, say, then each point $x$ has a compact neighbourhood $\{x,0\}$ but the boundaries of open sets are mostly not compact.