Discrete Topology, Separable iff Countable Let $X$ be a topology space with the discrete topology.
I want to show that $X$ is separable iff $X$ is countable. The countable $\Rightarrow$ separable part was easy, but the other way I'm not sure if I'm correct.
I thought that, if we suppose that $X$ is separable and not-countable, we have that $\exists Y \subset X$ with $\overline{Y}=X$ ($\overline{Y} =$ closure of $Y$).
But for all $Y\subset X$, we have that $Y$ is closed. That means $\overline{Y}=Y$ (Right?). Then $X=Y$ and we have $Y$ not-countable, so we have a contradiction.
Can you confirm if my thought is correct? Thanks.

It's slightly nicer to avoid unneccessary proofs from contradictions: suppose $X$ is separable and discrete, so that there is a countable dense subset $D$ of $X$. As all sets are closed in the discrete topology $X=\overline{D}=D$ is countable.
