= Solution
\b[A <Hausdorff> compact <topology>.] The empty set omits zero and $\mathbb Z$ has finite complement. An arbitrary union omitting zero belongs to the family. If a union contains zero, one of its constituent sets contains zero and is cofinite, so the union is also cofinite. A finite intersection either omits zero or is an intersection of finitely many cofinite sets and is therefore cofinite. This proves all <topology> axioms.
For distinct nonzero integers $a,b$, the disjoint open singletons $\{a\},\{b\}$ separate them. To separate $0$ and $n\ne0$, use the disjoint <open sets> $\mathbb Z\setminus\{n\}$ and $\{n\}$. Thus this is a <Hausdorff space>. Every <open cover> has a member containing zero, which leaves only finitely many integers uncovered; choose one member for each of those points. This gives a finite subcover and proves <compactness>. The space is the <one-point compactification> of the <discrete space> $\mathbb Z\setminus\{0\}$, with zero as the added point.
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