Weaker condition for compact sets to be closed

Question

It's a very standard topological proof that in Hausdorff spaces compact sets are closed. But is Hausdorff condition necessary, is there an example of topological space $X$ so that all compact sets are closed but space is not Hausdorff? If yes, is there a weaker condition which is necessary for all compact sets to be closed?

Answer 1

In the classical paper between $T_1$ and $T_2$ Wilansky introduces the notion of a KC space:

$X$ is called a KC space if every compact subset of $X$ is closed in $X$.

And as you note in the question, A Hausdorff ($T_2$) space is KC. And a KC space is clearly $T_1$ as all finite subsets are always compact and thus closed in a KC space.

The notion of US (unique sequential limits) is also introduced:

$X$ is US iff for all sequences $(x_n)$ from $X$, if $x_n \to x$ and $x_n \to x'$ then $x=x'$.

And again $T_2 \implies US \implies T_1$ (for the latter consider constant sequences).

A KC space that is not Hausdorff is $\alpha\Bbb Q$, the Alexandroff extension (one-point compactification) of the rationals.

When such spaces are compact they turn out to be the maximally compact spaces, see this answer, e.g. They're of some interest (look in Google scholar etc. for papers on them).