Past exam of the options I course of the University of Oxford / 2009 / E / E1 / b / i / Solution

Both $\varnothing$ and $X$ satisfy the condition. For an arbitrary intersection of members, either every member contains $a$, in which case the intersection contains $a$, or one does not, in which case that member is countable and so is the intersection. The empty intersection is $X$.

For a finite union, either a member contains $a$, so the union does too, or all members omit $a$ and are countable, so their finite union is countable. Thus the closed set axioms hold and define a topology. This is a Fortissimo space: every point other than $a$ is isolated, while the open neighbourhoods of $a$ have countable complement.