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Can someone verify my proof of the claim in the title?

Let $X$ be a Hausdorff space satisfyin the hypothesis in the title. For $x\in X$, there exists $K$ a compact neighborhood of it. Then $K$ is compact and Hausdorff, therefore $K$ is locally compact. For $O$ an open neighbourhood of $x$, $O \cap K$ is an open nbhd of $x$ in $K$. So by local compactness there exists a nbhd $L$ of $x$ in $K$ such that $L \subset O\cap K$ and $L$ is compact. Now let's show that $L$ is a nbhd of $x$ in $X$. Since $L$ is a nbhd of $x$ in $K$, there exists $A$ an open nbhd of $x$ in $K$ such that $A\subset L$. Write $A =A'\cap K$ for $A'$ open in $X$. Finally let $A'' = A' \cap K^{\circ}$. We know $K$ is a nbhd of $x$ so there exists $B$ open in $X$ such that $x \in B \subset K$. Then $K^{\circ} \supset B \ni x$. And $x \in A \implies x \in A'$. Thus $A''$ is an open nbhd of $x$ in $X$ and $A'' \subset A' \cap K =A \subset L$. Therefore $L$ is a nbhd of $x$ in $X$, $L$ is compact and $L \subset O$. Therefore $x$ has a compact local base. QED

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