TITLE: Subspace of metrizable and separable space is separable
QUESTION [6 upvotes]: I need to show (using the fact that for metrizable space: space is separable $\iff$ it has got a countable base) that if $X$ is metrizable and separable, then every subspace $Y \subset X $ is separable.
My idea:
If $X$ is separable (and is metrizable), then $X$ has got a countable base. Let's call it $\mathcal{B}$. Let $U=\bigcup_{i \in I} Bi$ such that $B_i \in \mathcal{B}$. So $U$ is open in $X$. Then $U\cap Y$ is open in $Y$. So $U \cap Y=\bigcup_{i \in I} Bi \cap Y$ such that $B_i \in \mathcal{B}$. So I want to say that $\mathcal{C}=\{B \cap Y: B\in\mathcal{B}\}$ is the base of $Y$. And becasue $\mathcal{B}$ is countable, then $\mathcal{C}$ is also countable. Then $Y$ is also separable.
Is it a correct proof (if not how can I improve it)? How should I show that $\mathcal{C}$ is really a base?

REPLY [3 votes]: Your argument is basically correct. You can state it a little more clearly by rearranging it a bit:

If $X$ is separable and metrizable, then $X$ has a countable base $\mathscr{B}$. Let $\mathscr{B}_Y=\{B\cap Y:B\in\mathscr{B}\}$; clearly $\mathscr{B}_Y$ is countable. Let $U$ be an open set in $Y$. Then there is an open set $V$ in $X$ such that $U=V\cap Y$. $\mathscr{B}$ is a base for $X$, so there is some $\mathscr{U}\subseteq\mathscr{B}$ such that $V=\bigcup\mathscr{U}$. Then $$U=V\cap Y=\left(\bigcup\mathscr{U}\right)\cap Y=\bigcup\{B\cap Y:B\in\mathscr{U}\}$$ is a union of elements of $\mathscr{B}_Y$, so $\mathscr{B}_Y$ is a base for $Y$. $Y$ is therefore second countable and hence separable.