As in this question, let $\tau$ be the standard topology on $\mathbb{R}$ and let $\tau'$ be the compact complement topology on $\mathbb{R}$: $$\tau' = \{\mathbb{R} \setminus K \mid K \text{ is $\tau$-compact}\} \cup \{\varnothing, \mathbb{R}\}$$
One of the commenters makes the claim that the $\tau'$-compact subsets are precisely the $\tau$-closed subsets, but I'm having trouble filling in the details. Here's what I have so far:
Suppose $A$ is $\tau$-closed and $\mathcal{U}$ is a $\tau'$-open cover of $A$. If $\mathbb{R} \in \mathcal{U}$, then $\{\mathbb{R}\}$ is a finite subcover. Otherwise, choose any nonempty $U \in \mathcal{U}$. By the definition of $\tau'$, $\mathbb{R} \setminus U$ is $\tau$-compact, so $A \setminus U=A \cap (\mathbb{R} \setminus U)$ is $\tau$-compact. Since $\tau' \subseteq \tau$, we have $\mathcal{U}$ is a $\tau$-open cover of $A \setminus U$. Let $\mathcal{V} \subseteq \mathcal{\mathcal{U}}$ be a finite subcover of $A \setminus U$. Then $\mathcal{V} \cup \{U\}$ is a finite subcover of $A$, so $A$ is $\tau'$-compact.
I don't have a proof of the converse. Given a $\tau'$-compact set $A$, we want to show that it is $\tau$-closed. The comment suggests supposing it is not $\tau$-closed. Then you could find a ($\tau$- ?) compact set $K$ such that the intersection $A \cap K$ is not ($\tau$- ?) closed (how?). We could then conclude that $A \cap K$ is $\tau'$-compact, but I don't see how that helps us.
Have I made any errors? Can anyone finish the proof or give a counterexample to the converse?