Definitions.
Let $X$ be a topological space. We say $X$ is
locally compact if every member of $X$ has neighborhood basis of compact sets.
weak Hausdorff if for every continuous map $f\colon Z\to X$, where $Z$ is compact Hausdorff, $f(Z)$ is closed in $X$.
Question.
($Q$) Must a locally compact weak Hausdorff space be Hausdorff?
Additional background.
This question has an equivalent formulation in terms of compactly generated spaces, or "$k$-spaces". We need a few more definitions.
A subset $A\subseteq X$ is $k_2$-closed (resp. $k_2$-open), if $f^{-1}(A)$ is closed (resp. open) for every continuous map $f\colon Z\to X$ from a compact Hausdorff space $Z$.
(Often the above is simply called "$k$-closed", but there are several competing definitions for that term, one of which we are about to introduce here, so we borrow the terminology from this answer for precision.)
A subset $A\subseteq X$ is $k_3$-closed (resp. $k_3$-open) if its intersection $A\cap K$ with every compact Hausdorff subset $K\subseteq X$ is closed (resp. open) in $K$.
For $n=2$ or $3$, $X$ is
$CG_n$ ("compactly generated") if subsets in $X$ are closed if and only if they are $k_n$-closed.
$k_n$-Hausdorff if the diagonal in $X\times X$ is $k_n$-closed.
The last definition really isn't especially interesting for $n=3$ in light of the answers here. Also there is a definition for $k_1$ as well; it won't be relevant for us, but see the preceding links.
Equivalent and related questions.
As discussed in the first link above, under either $CG_n$ assumption, the weak Hausdorff condition is equivalent to the otherwise stronger condition $KC$ (Compact subsets are closed). However, the $KC$ condition in turn immediately upgrades to $T_2$ in the presence of local compactness, as we obtain a basis of closed neighborhoods at each point, which combines with $T_1$ to yield $T_2$. (We get $T_1$ from the weak Hausdorff condition, as every singleton is the image of a compact Hausdorff space, hence closed.)
Conversely, any locally compact Hausdorff space is easily seen to be $CG_n$ for either $n$.
As a result, the following question, for either $n=2$ or $3$, is equivalent to ($Q$).
($Q_n$) Must a locally compact weak Hausdorff space be $CG_n$?
Finally, we note that in general, weak Hausdorff implies $k_2$-Hausdorff, and under either $CG_n$ assumption, the two are equivalent.
Therefore, we could also ask
($Q'$) Must a locally compact $k_2$-Hausdorff space be Hausdorff?
($Q_n'$) Must a locally compact $k_2$-Hausdorff space be $CG_n$?
An affirmative answer to either of these equivalent (to each other) questions implies an affirmative answer to the other questions.
In general, a big part of what motivated this question was my desire to find more examples of spaces that are $k_2$-Hausdorff or even weak-Hausdorff, yet fail to be $KC$. As explained above, in the locally compact case any non-Hausdorff example will also fail $KC$, leading me to this line of inquiry.
Addendum.
Regarding ($Q_n$), I had difficulty finding any example at all of a locally compact space that failed to be a $CG_2$-space. $\pi$-base turned up nothing, though this answer did give a nice non-$T_1$ example. (The original paper referenced there explains how to modify the example to be $T_1$, but after such modification it will still fail to have any reasonable separation, as it will not even have unique limits for convergent sequences ($US$), so will not answer our questions here.)
For locally compact spaces that fail to be $CG_3$, $\pi$-base only has non-$T_1$ examples.
