TITLE: Proof help. Core-compactness, Hausdorff, Locally Compact
QUESTION [5 upvotes]: 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.
EDIT
Here are the two papers which make make me believe this is a theorem:
Core Compactness and Diagonality in Spaces of Open Sets
Topologies on Spaces of Continuous Functions
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).
Non-Hausdorff Topology
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.
EDIT
I am now asking for help in completing the proof. This has been bothering me for too long.

REPLY [4 votes]: This can be done easily with ultrafilters. Similarly to the ultrafilter characterisation of compactness, $V\ll U$ is equivalent to saying that each ultrafilter containing $V$ has a limit in $U$. 
For the proof, let $x\in X$. By core-compactness, we can choose open neighborhoods $U$ and $V$ of $x$ such that $V\ll U$ and $U\ll X$. All ultrafilters containing $V$ have no limit outside of $U$, since they already have a limit in $U$, and they can only have one limit thanks to Hausdorffness. But this means that $\overline{V}\subseteq U$. Hence every ultrafilter containing $\overline{V}$ contains $U$, and since $U\ll X$, these ultrafilters converge. As $\overline{V}$ is closed, they converge to something in $\overline{V}$. So $\overline{V}$ is compact. So $x$ has a compact neighborhood.