An $\omega$-cover $\mathscr U$ of a space $X$ is a collection of open sets so that $X \not\in\mathscr U$ and every finite subset of $X$ is contained in a member of $\mathscr U$. Similarly, a $k$-cover $\mathscr U$ of a space $X$ is a collection of open sets so that $X \not\in\mathscr U$ and every compact subset of $X$ is contained in a member of $\mathscr U$.
A space is an $\varepsilon$-space (aka $\omega$-Lindelöf) if every $\omega$-cover has a countable subset which is an $\omega$-cover. Likewise, a space is $k$-Lindelöf if every $k$-cover has a countable subset which is a $k$-cover.
In an effort to establish the relationships between these properties, one may wonder if there are any implications either way. For example,
Is every $k$-Lindelöf space an $\varepsilon$-space?
As an initial attempt, one may start with an $\omega$-cover of $X$ and close it under finite unions to guarantee a $k$-cover of $X$. However, the countable subset which is a $k$-cover of the resulting $k$-cover may not correspond to an $\omega$-subcover of the given $\omega$-cover of $X$. So this argument doesn't work.
Edit: One approach occurred to me as I was writing the question. My posting here is to have a reference to include for https://topology.pi-base.org/ following the guidelines for references as spelled out at https://github.com/pi-base/data/blob/master/CONTRIBUTING.md
If anyone knows of a place this is mentioned in the literature that I missed, I'd appreciate knowing where it occurs.
Yes, every $k$-Lindelöf space is a $\varepsilon$-space. The proof follows the same line of reasoning as the proof of Theorem 5 in https://doi.org/10.1007/s10114-005-0753-8.
What we actually show is that every finite power of a $k$-Lindelöf space $X$ is $k$-Lindelöf and, since every $k$-Lindelöf space is Lindelöf, we see that any $k$-Lindelöf space is Lindelöf in each of its finite powers. (The fact that every $k$-Lindelöf space is Lindelöf is stated without proof in the proof of Proposition 5 of https://doi.org/10.1016/j.topol.2005.07.015, and is pretty straight-forward. Just take an open cover and consider the cover formed by the finite unions from that open cover. The result is a $k$-cover so you can choose a countable subset which is a $k$-cover. Note that this corresponds to a countable subcover of $\mathscr U$.) The fact that being an $\varepsilon$-space is equivalent to being Lindelöf in each of its finite powers is a well-known result from https://doi.org/10.1016/0166-8641(82)90065-7.
Proof. Let $X$ be $k$-Lindelöf and suppose $\mathscr U$ is a $k$-cover of $X^m$ for a positive integer $m$. By Lemma 3 of https://doi.org/10.1007/s10114-005-0753-8, we can let $\mathscr V$ be a $k$-cover of $X$ so that $\\{ V^m : V \in \mathscr V \\}$ refines $\mathscr U$. Then there is some $\\{V_n : n \in \omega\\} \subseteq \mathscr V$ that forms a $k$-cover of $X$. Let $U_n \in \mathscr U$, be so that $V_n^m \subseteq U_n$. To see that $\\{ U_n : n \in \omega \\}$ is a $k$-cover of $X^m$, let $K \subseteq X^m$ be compact. Note that the union of $\pi_{k}[K]$ for $1 \leq k \leq m$ is a compact subset of $X$ where $\pi_{k}$ is the usual projection mapping onto the $k^{\text{th}}$ coordinate. So we can find $n \in \omega$ so that $$\bigcup_{j=1}^m \pi_k[K] \subseteq V_n.$$ Hence, $K \subseteq V_n^m \subseteq U_n$. $\square$