Complementary compactness

Question

Let $X$ be a topological space having the property that whenever a subset $A$ of $X$ is compact, then $X\setminus A$ is compact too. Is every subset of $X$ compact?

Answer 1

Yes. Since $\emptyset$ is compact, $X$ is compact, so every closed subset of $X$ is compact, so every open subset of $X$ is compact. If every open subset of $X$ is compact, then every subset of $X$ is compact. That is, every family of open sets in $X$ has a finite subfamily with the same union.

Answer 2

$X$ itself is compact, hence Hausdorff. Singletons are compact, hence their complement is compact and closed, which makes singletons open and $X$ discrete. But then $X$ must be a finite discrete space, whence the claim.