Weakly Locally Compact Hausdorff Space is Strongly Locally Compact

Theorem
Let $T = \left({X, \vartheta}\right)$ be a $T_2$ (Hausdorff) space.

Let $T$ be weakly locally compact.

Then $T$ is strongly locally compact.

Proof
Let $x \in X$.

As $T$ is weakly locally compact, $x$ is contained in a compact neighborhood $N_x$.

As $T$ is a $T_2$ (Hausdorff) space, we can use the result Compact Subspace of Hausdorff Space is Closed.

Thus the interior of $N_x$ has a closure which is compact.

Hence the result, from definition of strongly locally compact space.

Remark
The assumption that $T$ is Hausdorff is necessary, as the following example shows:

Let $X$ be a infinite set. Let $\tau_p$ be the particular point topology, that is: For a $p \in X$ a non-empty set is open if it contains $p$. Also of course the empty-set is open.

$X$ is not compact: Take the covering $\{ \{ x, p \} \}_{x \in X}$. This has no finite sub cover.

As the closure of any open set is the whole space $X$, we have that $X$ is not strongly locally compact.

But then again $\{ x, p \}$ is compact, as it is finite. Hence $X$ is weakly locally compact.