Show that a dense subspace $Y$ of a first countable separable topological space $X$ is separable.
Proof:
$X$ is separable.
Let $S=\{x_n \in X | n \in \mathbb{N}\}$ be a countable dense subset of $X$.
$Y$ is also dense in $X$.
Because $X$ is first-countable, thus for each $x_n$ where $n \in \mathbb{N}$ there exists a countable local-basis around $x_n$.
Let the countable local-basis around $x_n$ be $S_n=\{\text{ }B_n^k \text{ } | \text{ }k \in \mathbb{N} \}$
Because $Y$ is dense in $X$ thus for each $x_n$ where $ n=1,2,3 \dots $ and for each $B_n^k$ where $k=1,2,3,4 \dots$, we have $Y \cap B_n^k \neq \phi$.
Say $y_n^k \in Y \cap B_n^k \neq \phi$
Denote $Z=\{ y_n^k \in Y \text{ } | \text{ } n,k \in \mathbb{N} \}$
Claim: $Z$ is a countable dense set of $Y$.
Choose $y \in Y$ and any open set $V$ in $Y$ containing y.
$V$ is open in $Y$ implies that $V=U \cap Y$ where $U$ is an open set in $X$.
Thus $y \in U \in \tau$ and $y \in Y$
$y \in U$ and $U$ is open in X. Because $S$ is dense in X, we have that $U \cap S \neq \phi $.
Let $x_n \in U \cap S$, Thus $x_n \in U$ and $U$ is open in $X$.
Considering that $S_n$ is a countable local-basis around $x_n$ we have an element $B_n^{k_0}$ such that $x_n \in B_n^{k_0} \subset U$. choose the corresponding $y_n^{k_0}$ as done in the construction above. Then we have $y_n^{k_0} \in B_n^{k_0} \subset U$. Thus $y_n^{k_0} \in U \cap Y = V$ and hence $V \cap Z \neq \phi$ as it contains $y_n^{k_0}$.
Hence $Y$ has a countable dense subset. $Y$ is separable.
Hence proved!
Please check my solution. I need to correct my mistakes and learn.
Thank You.
Best Answer
This proof looks fine. Quite detailed. See Daniel's comment for an alternative faster proof.
To see that you need the first countable assumption on $X$: if $X=[0,1]^\mathbb{R}$, then $X$ is separable (but not first countable), and $Y=\Sigma_0[0,1]^\mathbb{R} := |\{f \in X: |\{x: f(x) \neq 0\}| \le \aleph_0 \}$ is dense in $X$ and not separable. Think about it.