I am trying to prove that any subset $Y$ of a separable metric space $X$ is separable (in the induced metric). Is the following correct?
Let $\{z_{n}\}_{n \in \mathbb{N}}$ be the countably infinite subset of $X$ that is dense in $X$. I show that, for all $x \in Y$ and $\epsilon>0$, $B_{\epsilon}(x) \cap Y \cap \{z_{n}\}_{n\in \mathbb{N}}$ is non-empty (this trivially implies the theorem being proved).
Suppose, for a contradiction, that there exists $x \in Y$ and $\epsilon>0$ such that $B_{\epsilon}(x) \cap \{z_{n}\}_{n\in \mathbb{N}}$ \subset $X \setminus Y$. Note, then that $B_{\frac{\epsilon}{n}}(x) \cap \{z_{n}\}_{n\in \mathbb{N}} \subset Y$, for all $n \in \mathbb{N}$ (as otherwise there would exist a subset of $B_{\epsilon}(x) \cap \{z_{n}\}_{n\in \mathbb{N}}$ that is not a strict subset of $X \setminus Y$). This implies that infinitely many terms of $\{z_{n}\}_{n\in \mathbb{N}}$ are contained in $X \setminus Y$ (that is, outside of $Y$).
Note now that there cannot exist another $z \in Y$ and $r>0$, $z \neq x$, such that $B_{r}(z) \cap \{z_{n}\} \subset X \setminus Y$ (as otherwise, by the same reasoning as above, there would exist two countably infinite sequences contained inside the countably infinite sequence $\{z_{n}\}$—a contradiction). So then, choose an arbitrary $z \in Y$ and $r>0$. We must have that $B_{\frac{r}{n}}(z) \cap \{z_{n}\} \cap Y $ is non-empty for each $n \in \mathbb{N}$. This contradicts the fact that there are only finitely many terms of $\{z_{n}\}$ in $Y$.
Thank you.
Your contradiction is unfortunately not a contradiction. It is possible to extract two, or even countably many, disjoint subsequences from a countably infinite sequence.
Your goal of proving that $Y$ has nonempty intersection of $(z_n)$ is also doomed, because $Y$ can be entirely disjoint from $(z_n)$. Consider $X={\mathbb R}$, $(z_n)$ an enumeration of the rational numbers, and $Y$ the set of irrational numbers.