I'm trying to prove the following statement:
Every subspace of a separable metric space is separable.
Could you please verify if I correctly apply the concept relative topology? Thank you so much!
$\textbf{My attempt}$
First, I need the following lemma:
Let $X$ be a metric space. The following statements are equivalent:
$X$ satisfies the second countability axiom.
$X$ is a Lindelöf space.
$X$ is separable.
Let $(X,d)$ be a separable metric space, $Y \subseteq X$, and $d_Y$ the induced metric of $d$ on $Y$. By Lemma, $X$ satisfies the second countability axiom. Then the topology induced by $d$ has a countable basis $\mathcal M$. Let $\mathcal N = \{A \cap Y \mid A \in \mathcal M\}$. By relative topology, $\mathcal N$ is a countable basis of the topology induced by $d_Y$. Hence $(Y,d_Y)$ satisfies the second countability axiom. Thus $(Y,d_Y)$ is separable by Lemma.
Update: From @Henno Brandsma's answer, I include the proof that $\mathcal N$ is a basis of the topology induced by $d_Y$.
Assume $A \subseteq Y$ is open in $Y$. Then $A = O \cap Y$ for some $O \subseteq X$ that is open in $X$. It follows from $O$ is open in $X$ that there is $\mathcal M' \subseteq \mathcal M$ such that $O = \bigcup \mathcal M'$. Let $\mathcal N' = \{Y \cap B \mid B \in \mathcal M'\}$ Then $\mathcal N' \subseteq \mathcal N$ and $A = \bigcup \mathcal N'$. Hence $\mathcal N$ is a basis of the topology induced by $d_Y$.
Best Answer
In this answer I show a bigger equivalence, with all proofs.
The fact that second countable implies that all subspaces are separable you have shown correctly. You need a small lemma that when $\mathcal{B}$ is a base for $X$, its intersections with $Y$ form a base for $Y$ too, but I assume you already showed that before.