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

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.

Let $S=\{x:f(x)\ne f(a)\}$, which is countable. For any closed set $C\subseteq \mathbb{R}$, consider its preimage. If $f(a)\in C$, then $a\in f^{-1}(C)$ and that preimage is closed by definition. If $f(a)\notin C$, then $f^{-1}(C)\subseteq S$, so it is countable and again closed. Therefore every closed set has a closed preimage, proving continuity.

Set $S=\{x_n:n\in \mathbb{N},x_n\ne \ell \}$. This is a countable set, hence a closed set in the stated topology, and $\ell \notin S$. Consequently $X\setminus S$ is an open neighbourhood of $\ell$. By convergence of a sequence, all sufficiently late terms lie outside $S$. A sequence term outside $S$ must equal $\ell$. Thus the sequence is eventually constant, and only finitely many indices have $x_n\ne \ell$.