Two (maybe nonequivalent) definitions of local compactness

Question

1. $X$ is locally compact if every point has a neighborhood with a compact closure.
2. $X$ is locally compact if every point lies in the interior of a compact subspace of $X$.

Clearly, $(1) \implies (2)$.

I am wondering whether $(2) \implies (1)$ or not. (When $X$ is not necessarily Hausdorff)

Can you give a proof/counterexample for this?

Thanks in advance.

Answer 1

Let $X$ be an infinite set, $p$ a point of $X$, and $\tau=\{\varnothing\}\cup\{U\subseteq X:p\in U\}$; $\tau$ is a topology on $X$ that satisfies $(2)$, since $x\in\operatorname{int}\{x,p\}$ for each $x\in X$. However, it fails to satisfy $(1)$, since the only closed set containing $p$ is $X$ itself, which is not compact: $\big\{\{x,p\}:x\in X\big\}$ is an open cover with no finite subcover.