Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 12F / 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 $n\ne 0$, 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.