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 $\mathbb{F}$ be an open cover of $S$ Let $F \in \mathbb{F}$ Then $X\setminus F$ is finite. Choosing a member of $\mathbb{F}$ for each remaining element of $S$ shows that $\mathbb{F}$ 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.