The property that compact sets in a space are closed lies strictly between T1 and T2

If a topological space is Hausdorff ($T_2$), then every compact subset of that space is closed. If every compact subset of a space is closed, then (since singletons are always compact) then the space is accessible ($T_1$). There are spaces that are $T_1$ and have compact sets that are not closed, and there are spaces in which compact sets are always closed but that are not $T_2$.

Let $X$ be an infinite set with the finite complement topology. Singletons are finite, and therefore closed, so $X$ is $T_1$. Let $S\subset X$, and let $𝔽$ be an open cover of $S$. Let $F\in 𝔽$. Then $X\setminus F$ is finite. Choosing a member of $𝔽$ for each remaining element of $S$ shows that $𝔽$ has a finite subcover. Thus, every subset of $X$ is compact. An infinite subset of $X$ will then be compact, but not closed.

Let $Y$ be an uncountable set with the countable complement topology. No two open sets are disjoint, so $Y$ is not Hausdorff. Let $C$ be a compact subset of $Y$. I shall show that $C$ is finite. Suppose $C$ is infinite, and let $S$ be an infinite sequence in $C$ without any repetitions. For any natural number $n$, let $U_n$ be all the elements of $C$ except for all the $S_k$, where $k>n$. Then $U_n$ is open for each $n$, and $\{U_n\mid n\in \mathbb{N}\}$ covers $C$, but has no finite subset that covers $C$, contradicting the fact that $C$ is compact. This contradiction arose by assuming a compact subset of $Y$ was infinite, all compact subsets of $Y$ are finite. $Y$ is $T_1$ (singleton sets are countable), so all compact subsets of $Y$ are closed.

These examples were suggested by the person known as Polytope on EFNet’s math channel.