Does locally compact and $\sigma$-compact non-Hausdorff space imply hemicompact? I know $\sigma$-compact locally compact Hausdorff spaces are hemicompact. What if we remove the Hausdorff condition or replace Hausdorff with $T_1$?
 A: Yes, you can drop the Hausdorff condition completely. In analogy to the theorem, that locally compact and $\sigma$-compact spaces are paracompact (See here), we have:
Theorem: Locally compact and $\sigma$-compact spaces are hemicompact (Proof see here).
There is an example of a hemicompact, but not locally compact space found in Example 10, Chapter 4, Section 2 in Introduction to General Topology by K. D. Joshi. Nonetheless, a first countable and hemicompact space is locally compact (Proof see here). The inverse direction holds for the other condition though: Hemicompact spaces are $\sigma$-compact (See here). On the other hand, $\mathbb{Q}$ is a $\sigma$-compact, but not hemicompact space (See here).
A: As you probably know, without the Hausdorff property there are various inequivalent notions of locally compact space.  The weakest of them being weakly locally compact, that is every point has a compact nbhd (= condition (1) in Wikipedia).
Now the planetmath proof of the result mentioned in @SamuelAdrianAntz's answer seems to make some assumptions (open sets with compact closure, ...) which do not always hold.  But the following more general result is true.

Theorem: Every weakly locally compact $\sigma$-compact space is hemicompact.

Proof: As shown in this answer, being weakly locally compact and $\sigma$-compact is equivalent to being exhaustible by compact sets.  So we can find a sequence of compact sets $K_n$ in the space $X$, each contained in the interior of the next one, and whose union is the whole set.  Note that the interiors of all the $K_n$ also cover the whole space.  Now any given compact set $K\subseteq X$, it is contained in the union of the interiors of the $K_n$.  Thus, by compactness it is contained in the interior of one of the $K_n$, and also in $K_n$.  In other words, $X$ is hemicompact.
