Counterexample: locally compact non-Hausdorff space where open sets are sigma-compact and compact sets are closed?

Question

Question. Does there exist a topological space $(X, \tau)$ which simultaneously satisfies all four of the following criteria?

1. It is locally compact, meaning every point $x \in X$ has a compact neighbourhood.
2. Every compact set is closed.
3. Every open set $\mathcal{U} \in \tau$ is σ-compact, i.e., can be written as an at most countable union of compact sets.
4. It is non-Hausdorff.

No finite counterexample can exist. For then all subsets are compact, therefore closed, yielding the metrisable discrete topology.

The cocountable topology on an uncountable set would satisfy 2. and 4., but neither 1. nor 3. hold. For example, 3) fails because precisely finite sets are compact which makes writing the uncountable $X$ unfeasible with anything other than an uncountable union.

Any ideas?

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The question here is self-posed. It comes from an analysis of a proof in introductory measure theory. A certain regularity result with Hausdorffness as an assumption seems to only use the weaker assumption that all compact sets be closed. The details do not matter, but essentially the question is whether any generality would be gained by switching Hausdorfness with assumption 2.

Answer 1

The one-point compactification of $\Bbb Q$ (call it $\widehat{\Bbb Q}$) is compact (hence it has [1]) and countable (hence it has [3]). Since (closed) compact subsets of $\Bbb Q$ have empty interior, every neighbourhood of $\infty$ is dense, so $\widehat{\Bbb Q}$ isn't T2. According to pi-base, $\widehat{\Bbb Q}$ is KC (i.e. [2]). A proof of this fact is here.