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Every second countable space must be Lindelöf – given a cover, decompose each open set into basic open sets. This is a countable cover that refines the original cover, so by picking arbitrary open sets from the original cover containing each of the countably-many basic open sets in our refinement, we obtain a countable subcover.

A space is $k$-Lindelöf if every $k$-cover (each compact is contained in a single open set) admits a countable $k$-subcover.

All separable metrizable spaces must be second countable and must be $k$-Lindelöf (due to a theorem later deprecated in favor of this based on this post). There are several spaces that are $k$-Lindelöf but not second countable. The pi-Base does not know any second countable space that's not $k$-Lindelöf.

Is every second countable space in fact $k$-Lindelöf?

Steven Clontz
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Yes. Let $\mathcal B$ be a countable base for the topology of $X$ and let $\mathcal C$ be any open k-cover of $X$. Let $\mathcal U$ be the collection of all sets $U$ such that (i) $U$ is the union of finitely many elements of $\mathcal B$, and (ii) $U\subseteq C$ for some $C\in\mathcal C$. Then $\mathcal U$ is a countable open k-cover of $X$. For each set $U\in\mathcal U$ choose a set $C_U\in\mathcal C$ with $U\subseteq C_U$; then $\{C_U:U\in\mathcal U\}$ is a countable k-subcover of $\mathcal C$.

PatrickR
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bof
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    I needed to type this out: $\mathcal U$ is a k-cover because given a compact $K$, $K\subseteq C\in\mathcal C$. For each point $k\in K$ let $B_k$ be basic open with $k\in B_k\subseteq C$. Then there is a finite collection of $B_k$s covering $K$ whose union is some $U\in\mathcal U$, and thus $K\subseteq\mathcal U$. – Steven Clontz Jun 30 '23 at 04:43
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    Er, $K\subseteq U\in \mathcal U$. – Steven Clontz Jun 30 '23 at 14:32