While reading about topologies on continuous function spaces, I've seen remarks that core-compact and locally compact are equivalent for Hausdorff spaces.
Now I can clearly see that locally compact always implies core-compact, so the Hausdorff condition comes into the proof of the converse.
Let $K$ be a core-compact Hausdorff space. Let $x\in U\subseteq K$ with $U$ open. I need to show that $U$ contains a compact neighborhood of $x$. Since $K$ is core-compact there is an open neighborhood $V$ such that $x\in V\subseteq U$ with $V\ll U$.
I don't know where to go from here.
Here are the two papers which make make me believe this is a theorem:
I have found another source which claims something better: every sober core-compact space is locally compact (although I can't see the proof of the theorem; I might buy it).
This hints that the property of Hausdorff spaces that we want to exploit is the fact that the intersection of all closed neighborhoods of a point is precisely that point.
I am now asking for help in completing the proof. This has been bothering me for too long.