Characterizing $\sigma$-compactness via closed sets A topological space that is a countable union of compact subsets is called $\sigma$-compact.
My intuition says that the follow property should be equivalent to $\sigma$-compactness:
$$
\text{Every closed set is a countable union of compact sets}
$$
Is this actually the case?
Note: I'm thinking only of Hausdorff spaces here (it seems like this should fail in the non-Hausdorff case). 
 A: If $X = \bigcup_{i \in \mathbb{N}} K_i$ is $\sigma$-compact and Hausdorff (with the $K_i$ compact, and hence closed), then given any closed $F \subseteq X$ we have $F = \bigcup_{i \in \mathbb{N}} ( F \cap K_i )$ with each $F \cap K_i$ compact.
The other direction is trivial since $X$ is closed in itself.
A: Your intuition is correct, also if the space is not Hausdorff.
Let $X$ denote the underlying set of the topological space you are working in.
If every closed set is a countable union of compact sets then $X$ is a countable union of compact sets. This because $X$ is closed.
If conversely $X$ is a countable union of compact subsets and $F\subset X$ is closed then the intersections of these compact sets with $F$ are compact and $F$ equals the union of these intersections. Any intersection of a compact and a closed set is compact. You do not need Hausdorff for this:
Let $C$ be compact, $F$ closed and let $\mathcal{U}$ denote an
open cover of $C\cap F$. Then $\mathcal{U}\cup\left\{ F^{c}\right\} $
is an open cover of compact $C$. It has a finite subcover $\mathcal{U}'$
and $\mathcal{U}'\backslash\left\{ F^{c}\right\} \subset\mathcal{U}$
is a finite subcover of $C\cap F$.
