Without Hausdorff, what implications can we prove about $k_\omega$ related to other covering properties? In e.g. A SURVEY OF $k_\omega$-SPACES a space is said to be $k_\omega$ if it's the union of compact Hausdorff $K_n$, $n<\omega$, with a set being closed if and only if its intersection with each $K_n$ is closed.
It's asserted there that $k_\omega$ is implied by $K_n\subseteq int(K_{n+1})$, i.e. exhaustible by (Hausdorff) compacts.
I want to consider the case where the $K_n$ need not be Hausdorff. In this case, we have exhaustible by compacts $\Rightarrow$ hemicompact $\Rightarrow$ $\sigma$-compact (with no arrows reversing). Where does this non-$T_2$ $k_\omega$ live?

Exhaustible by compacts implies this $k_\omega$. Let $K_n\subseteq int(K_{n+1})$. $k_\omega$ is equivalent to a set having open intersection with each $K_n$ with respect to the subspace topology implies the set is open. So let $U$ have open intersection with each $K_n$ and let $x\in U$. Note $x\in K_n$ for some $n<\omega$. Then let $V$ be an open set such that $V\cap K_{n+1}=U\cap K_{n+1}$. $x\in V\cap int(K_{n+1})\subseteq U$, proving $U$ is open.
Obviously, $k_\omega$ still implies $\sigma$-compact. Where does hemicompact fit in without something like Hausdorff or locally compact?
 A: Edit: My original post contained a faulty counterexample with a rather obvious flaw. Since the rest of the post contained some useful (but well-known) material, I will leave this up as an extended comment.
So you want to consider a space $X$ which is the colimit of an expanding sequence
$$X_1\subseteq X_2\subseteq\dots\subseteq X_n\subseteq\dots$$
of its compact (not necessarily Hausdorff) subspaces? Such an $X$ is Lindelöf, and is $T_1$ when each $X_n$ is $T_1$.

In case $X$ is $T_1$, it is hemicompact.

Proof: Suppose that $K\subseteq X$ is a compact subset not contained in any $X_n$. We can assume without loss of generality that the $X_n$ are all distinct and that $K\cap X_1\neq\emptyset$. For each $n\geq0$ choose $x_n\in K\cap(X_n\setminus X_{n-1})$, where we understand $X_{0}=\emptyset$. Write $D=\{x_n\mid n\in\mathbb{N}\}\subseteq X$. If $C\subseteq D$ is any subset, then $C$ is closed in $D$, since for each $n$ we have that $C\cap X_n$ is finite and hence closed in $X_n$. It follows that $D$ is infinite, discrete and closed in $X$, and this contradicts the compactness of $K$. $\;\blacksquare$
On the other hand, with not separation assumptions we have;

If $X_n$ is open in $X_{n+1}$ for each $n\in\mathbb{N}$, then $X$ is hemicompact.

Proof: The family $\{X_n\}_{n\in\mathbb{N}}$ is a an open cover of $X$. $\;\blacksquare$
Example 0: A hemicompact example which is not exhaustible by compacts.
For $n\in\mathbb{N}$ let $X_n=\bigvee^n_{i=1}S^1$ and equip $X=\bigcup^\infty_{i=1}X_n$ with the colimit topology. As a countable CW complex, $X$ is a Hausdorff $k_\omega$-space. If $X$ had a sequence of exhaustion $K_1\subseteq int(K_2)\subseteq K_2\dots$, then each point of $X$ would be contained in the interior of one of the compact $K_n$'s. Since each compact subset of a CW complex is metrisable, each point of $X$ would therefore have a metrisable neighourhood. But a locally metrisable paracompact space is metrisable, and $X$ is not even first-countable at the wedge point. $\;\square$
A: Not an answer, but some obervations.  Hereafter $k_\omega$ will mean the version without Hausdorff.
A hemicompact space need not be $k_\omega$, even in the Hausdorff case.  The survey article gives as an example on p. 114 a certain subspace of the Cech-Stone compactification $\beta\mathbb{N}$.
Another example is the Arens-Fort space.  As shows here this space $X$ is hemicompact.  But it is not $k_\omega$.  Indeed, it is anticompact and $T_1$, so every compact subspace is finite with the discrete topology.  So every subset $A\subseteq X$ meets every compact subspace $K\subseteq X$ in a closed subset of $K$, but not every subset of $X$ is closed since the space is not discrete.  This shows that $X$ does not admit any $k_\omega$ decompsition $(X_n)$.
So the other question is: does $k_\omega$ imply hemicompact?
The tricky thing seems that given a $k_\omega$ decomposition $(X_n)$, it is not necessarily the case that the space will be hemicompact with respect to that decomposition, but it could still be hemicompact with respect to a different decomposition.
Example 1 from Tyrone (now edited out, but available in the edit history) illustrates this.  Taking $X=\mathbb{N}=\{1,2,...\}$ with the indiscrete topology, the sets $X_n=[1,n]$ (in the usual order on $X$) form a $k_\omega$ decomposition but not every compact subset ($X$ in particular) is contained in one of the $X_n$.  But $X$ is still hemicompact because it is compact.
Another example is $X=\mathbb{Z}$ with the right order topology.  Open sets are of the form $[n,\infty)$ and compact sets are the ones with a minimum element.  The sets $X_n=[-n,n]$ form a $k_\omega$ decomposition (easy to check the weak topology condition), but the compact set $[0,\infty)$ is not contained in any of the $X_n$.  Note that in this case the closure of any of the $X_n$ is not even compact.  But $X$ is still hemicompact by taking $X_n=[-n,\infty)$ for example.

(Added 12/11/2022): The paper https://arxiv.org/abs/1602.04857 by T. Banakh uses the notions of $k_\omega$ and hemicompact spaces without assuming Hausdorff.
On the bottom of p. 8 it states:

"Each $k_\omega$-space is hemicompact."

Also on p. 16 before Lemma 2.5.1 it restates the same thing:

"It is known that each $k_\omega$-sequence $(K_n)_{n\in\omega}$ in a topological space X swallows compact subsets of the space
in the sense that each compact subset $K\subset X$ is contained in some set $K_n$."

But this last sentence cannot be true in general without Hausdorff, as my two examples above show.  Could be worth asking the author about it.
A: A partial result adapting 4.3 of this paper assuming the space is first-countable and $T_1$. Let $K_n$ for $n<\omega$ witness some $k_\omega$ decomposition with $K_n\subseteq K_{n+1}$. Suppose, aiming for a contradiction, some $p\in X$ has no compact neighborhood; in particular, no $K_n$ is a compact neighborhood of $p$. Without loss of generality, assume $p\in K_0$. Then if $\{U_n:n<\omega\}$ is the guaranteed (by first-countability) countable base at $p$, $U_n\not\subseteq K_n$; pick $p_n\in U_n\setminus K_n$. Then $\{p_n:n<\omega\}$ is closed as it has finite (a subset of $\{p_m:m<n\}$, and therefore closed by $T_1$) intersection with each $K_n$. But $p\not\in\{p_n:n<\omega\}$ is a limit point as $p_n\in U_n$ for all $n<\omega$, a contradiction.
So every first-countable $T_1$ $k_\omega$ space is weakly locally compact. Since weakly locally compact and $\sigma$-compact spaces are hemicompact, and first-countable hemicompact spaces are all weakly locally compact, $k_\omega$ and hemicompact are equivalent for first-countable $T_1$ spaces.
