TITLE: Compactness and Hausdorffness
QUESTION [6 upvotes]: Is it possible to have a topological space having all compact subsets closed, but the space itselt is not Hausdorff?
I couldn't find any counterexample. I tried $\mathbb R$ with cofinite, point inclusion and point exclusion topologies, but these didn't work.
Any suggestion in this context will be helpful for me.

REPLY [3 votes]: Another example of a $KC$ space is the real line with the cocountable topology, where the open sets are just the complements of countable sets. An infinite set $A$ contains a countable subset $a_1, a_2, ... $. The cover $U_1, U_2, ...$ where $U_n=\Bbb R-\{a_n,a_{n+1},...\}$, is then an open cover without a finite subcover. This means that $A$ is not compact. In other words, the compact sets in this topology are just the finite ones, so they are closed.
Note that a $KC$ space has unique sequential limits, i.e. a sequence that converges has a unique limit. In a first countable space it can then be shown that the space must be Hausdorff.