 Solution

The empty set is countable, and the whole space contains $a$, so both belong to the proposed closed family. A finite union contains $a$ if any member does; otherwise it is a finite union of countable sets and remains countable. For an arbitrary intersection, if every member contains $a$, so does the intersection. Otherwise some member excludes $a$ and is countable, and the intersection is a subset of it, hence countable. Thus all closed-set topology axioms hold.