I am trying to prove that if $X$ is a separable space, and $\{z_{n}\}$ is its countable dense subset, then every open subset of $X$ contains infinitely points from $\{z_{n}\}$. Is the following correct?
Consider an arbitrary open subset $Y \subset X$. Choose an arbitrary point in it, $z$. By the openness of $Y$, there exists $\epsilon>0$ such that $B_{\epsilon}(z)$ is wholly contained in $Y$. Also, by the separableness of $X$, $B_{\frac{\epsilon}{j}}(z)$ must contain at least one point from $\{z_{n}\}$, for each $j \in \mathbb{N}$. Cleary, this implies that $B_{\epsilon}(z)$, and so $Y$, contains infinitely many points from $\{z_{n}\}$.
Thank you.
No, it is not correct. For different values of $j$ you may get the same point of $(z_n)$ so there is no guarantee that you get infinitely many points of $(z_n)$.
The result is not even true.
Consider $\mathbb N$ with the (usual) discrete metric. $\mathbb N$ is itself a countable dense set and $\{1\}$ is an open set. It contains only one point from the countable dense set $\mathbb N$.