Crossposted to https://mathoverflow.net/questions/449504
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.
So the primary question is nothing beyond the title:
Is there a (Hausdorff) $\varepsilon$-space which is not $k$-Lindelöf?
To provide a little more context, Proposition 5 of https://core.ac.uk/reader/82129820 shows that a ($T_1$) space that is $k$-Menger and first countable is hemicompact where $k$-Menger is a selection principle that implies $k$-Lindelöf. However, in the proof, what they show is that a ($T_1$) space that is first countable but not locally compact is not $k$-Menger. The fact that they are using finitely many selections from each $k$-cover in a sequence of $k$-covers produces a convergent sequence of points; that sequence of points along with its limit is the compact set which cannot be covered. The obstacle in trying to adapt this argument to the $k$-Lindelöf scenario is that we can't guarantee such a selection against the countable base at a point which has no compact neighborhood.
As another comment, every $\sigma$-compact space is $\omega$-Menger and every $\omega$-Menger space is an $\varepsilon$-space (see https://people.math.wisc.edu/~awmille1/res/jmss.pdf). So the natural candidates are probably $\sigma$-compact spaces which are not hemicompact. The rationals are my favorite space of this class, so are the rationals $k$-Lindelöf?
Update: After a little more thought, the rationals are not the desired example. In fact, no separable metrizable space will work.
Claim: Every separable metrizable space is $k$-Lindelöf.
Proof. Suppose $X$ is a separable metric space. Then $\mathbb K(X)$, the hyperspace of non-empty compact subsets of $X$ with the Vietoris topology, is a separable metrizable space; thus $\mathbb K(X)$ is Lindelöf. By Corollary 4.16 of https://doi.org/10.1016/j.topol.2021.107772 (arXiv: [Cor. 31] https://arxiv.org/abs/2102.00296), $X$ is $k$-Lindelöf. $\square$
Best Answer
(MA) There exists a countable, hereditarily (strongly) paracompact space X, with ω<kL(X) (Example 6.4 in "Tightness, character and related properties of hyperspace topologies" Topology and its Applications 142 (2004) 245–292).