I attempted to find sufficient conditions for a space to be locally compact given that every point has a compact neighborhood, and I found that being regular is sufficient. I'm not sure my proof is correct so I will write it down. Thanks for your help.
Suppose that $(X,T)$ is a regular topological space in which every point has a compact neighborhood. Then $(X,T)$ is locally compact.
Let $x\in X$ and $U$ be a neighborhood of $x$. We know that there exists a compact neighborhood $K$ of $x$. We have to prove that there exists a compact neighborhood $V$ of $x$ such that $V \subset U$. Since $X$ is regular, there exists an open neighborhood $O$ of $x$ such that $\overline O \subset U$. Let $V = \overline O \cap K$. Then $V$ is a closed subset of $K$, so it is a compact subset of $K$, therefore $V$ is a compact subset of $X$. Since $K$ is a neighborhood of $x$, there exists an open neighborhood $O'$ of $x$ such that $O' \subset K$. Then $O\cap O'$ is an open neighborhood of $x$ and $O\cap O' \subset \overline O \cap K = V$, so $V$ is a neighborhood of $x$, it is compact, and $V \subset U$ since $\overline O \subset U$.