Past exam of the options I course of the University of Oxford / 2005 / E / 1 / Solution

A topology on $X$ contains $\varnothing ,X$, is closed under arbitrary unions and under finite intersections. The cofinite family has these properties: a nonempty union contains one cofinite member and therefore has finite complement; the complement of a finite intersection is a finite union of finite sets. The empty union and the sets $X,\varnothing$ are included explicitly.

Any two nonempty open sets intersect, because their intersection has finite complement in the infinite set $X$. Thus no two disjoint nonempty open sets can separate $X$, proving connectedness. For compactness, take any open cover and choose a nonempty member $U$. Only finitely many points lie outside $U$. Choose one covering member for each of those points. Together with $U$ these give a finite subcover. Thus $X$ is compact.