*Crossposted from https://math.stackexchange.com/questions/4717613*

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.

It can be shown that $k$-Lindelöf $\implies$ $\varepsilon$-space $\implies$ Lindelöf. The Sorgenfrey line is an example of a space which is Lindelöf but not an $\varepsilon$-space. This leads us to the following question.

Is there an $\varepsilon$-space which is not $k$-Lindelöf? Can the example be Hausdorff? Metrizable?

If it can be of any help, it would suffice to find a non-compact $\varepsilon$-space $X$ so that $\mathbb K(X)$, the space of compact subsets of $X$ with the Vietoris topology, is not Lindelöf by Cor. 4.16 of https://doi.org/10.1016/j.topol.2021.107772.

No such example exists in the realm of separable metrizable spaces.

Claim.Every separable metrizable space is $k$-Lindelöf.

*Proof.*
If $X$ is separable metrizable, then $\mathbb K(X)$ is also separable metrizable; hence, Lindelöf. If we let $[U] = \{ K \in \mathbb K(X) : K \subseteq U \}$, then, for a given $k$-cover $\mathscr U$ of $X$, $\mathscr W := \{ [U] : U \in \mathscr U \}$ is an open cover of $\mathbb K(X)$. Then a countable subcover of $\mathscr W$ corresponds to a countable subset of $\mathscr U$ which is a $k$-cover of $X$. $\square$