# If every countable subset of a space is closed, is the topology discrete?

I'm trying to solve Exercise 19, from Dugundji's Topology, pág. 92:

Let $X$ be a topological space in which every countable subset is closed. Is $X$ necessarily discrete?

I think the claim is false, but I couldn't figure out any counterexample.

Let's try an example. We'd better take an uncountable set; how about $\Bbb R$. Let's make all countable subsets closed, and make $\Bbb R$ closed. Now is there a topology on $\Bbb R$ making all these sets and no others closed? And is it a discrete topology?
• Thank you @LordSharktheUnknown. Ideed this defines a topology on $\Bbb{R}$, by one other exercise from Dugundji: "If card$(X)\geq \aleph_0$, then $\scr{A}_1$$=\{\emptyset\}\cup\{ A\,|\, \text{card}(X-A)<\text{card}(X)\}$ is a topology." $X=\Bbb{R}$ gives your example. And, indeed, this is not the discrete topology on $\Bbb{R}$, since $\{0\}$ is not open. – Ders Oct 13 '17 at 16:23