= Every subset of a cofinite space is compact
Every <subset> $A$ of a <cofinite> <topological space> is a <compact subset>. To see the finite reduction directly, take an <open cover> of a nonempty $A$ by <open sets> of $X$ and choose one member $U$ containing a point of $A$. Since $U$ has finite <complement>, the <set> $A\setminus U$ is finite. Choose a covering member for each of those remaining points. Those members together with $U$ are a <finite subcover>. The <empty set> is already <compact>.
Equivalently, the <subspace topology> on $A$ is itself cofinite: for finite $F\subseteq X$, the trace of $X\setminus F$ is $A\setminus(F\cap A)$. The argument applies again inside $A$. A concrete <open cover> $\{X\setminus\{k\}:k\in X\}$ of $X=\mathbb N_{>0}$ needs only the two members $X\setminus\{1\}$ and $X\setminus\{2\}$ when these labels are distinct: each fills the other's single gap.
On $X=\mathbb N_{>0}$, the <subset> of <even integers> is <compact> but not <closed>, because it is infinite and proper. A <compact subset of a Hausdorff space> is guaranteed to be <closed>, whereas <compactness> without the <Hausdorff space> assumption does not suffice. Changing the <topology> changes <compactness>: the same infinite <set> with its <discrete topology> has the <singleton> <open cover>, which has no <finite subcover>. The finite-exception argument is special to finite <complements>; in the <cocountable topology>, <compact subsets of a cocountable space> must instead be finite.
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