= Solution
A <Hausdorff space> is a <topological space> in which any two distinct points have disjoint open neighborhoods. Let $K$ be a <compact subset> of a <Hausdorff space> and $x\notin K$. For each $y\in K$, choose disjoint open neighborhoods $U_y$ of $y$ and $W_y$ of $x$. The $U_y$ cover $K$, so <compactness> gives a finite subcover $U_{y_1},\ldots,U_{y_m}$. Their paired intersection $W=\bigcap_{r=1}^mW_{y_r}$ is an open neighborhood of $x$ disjoint from $K$. Thus every point of its complement has an open neighborhood in the complement, proving \b[every <compact subset of a Hausdorff space> is closed]. The empty set case is immediate.
In the <cocountable topology> on an uncountable set, any two nonempty open sets intersect: their combined complements are countable and cannot exhaust $X$. Thus \b[the space is not <Hausdorff>].
Nevertheless \b[every <compact subset> is closed] here. In fact only finite subsets are compact. To prove this, suppose $K$ is infinite and choose distinct points $d_1,d_2,\ldots\in K$. The open sets
$$
U_n=X\setminus\{d_m:m\ge n\}
$$
form an increasing <open cover> of $X$, and hence of $K$. Any finite selection is contained in $U_N$ for its largest index $N$, which misses $d_N\in K$. So $K$ is not compact. Finite sets are compact in any <topological space>, and in this <cocountable topology> they are closed because their complements are open. This proves the claimed behavior of <compact subsets of a cocountable space>.
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