Continuous Map and Separable Subsets Suppose $X$ and $Y$ are metric spaces with $f: X \rightarrow Y$, where $f$ is continuous. I want to show that if $Q \subset X$ is separable, then $f(Q) \subset Y$ is separable.
My proof is as follows:
Let $f(X)=Z$. Then, we want to show that $Z$ is separable. First, we want to prove that for a continuous function $f$, $f(\bar{A})\subset \overline{f(A)}$
Suppose $B$ is some subset of $Z$. Then since $f$ is continuous and $\bar{B}$ is closed in $Z$, we know $f^{-1}(B)$ is closed.
Now, for a closed set $B$, $\bar{B}=B$. So $\overline{f^{-1}(\bar{B})}=f^{-1}(\bar{B})$.
We have that, $B\subset \bar{B} \implies f^{-1}(B) \subset f^{-1}(\bar{B})$
It follows that $\overline{f^{-1}(B)}\subset \overline{f^{-1}(\bar{B})}=f^{-1}(\bar{B})$
Set $B=f(A)$, and we get $\overline{f^{-1}(f(A))} \subset f^{-1}(\overline{f(A)})$ or, because $ff^{-1}=I$, $f(\bar{A}) \subset \overline{f(A)}$.
Now let $A$ be a countable dense subset of $X$ (this exists because $X$ is separable). Then $f(A)$ is clearly countable.
Because $A$ is dense in $X$, $\bar{A}=X$. So $Z=f(X)=f(\bar{A})$
So, we have $f(\bar{A}) \subset \overline{f(A)}$
So $Z \subset \overline{f(A)}$.
But, $\overline{f(A)} \subset Z$ as well.
Therefore $\overline{f(A)}=Z$
So we have found a countable set $f(A)$ which is dense in $Z$.
So $Z$ is separable.
 A: Choose a countable subset $Q_{1}\subseteq Q$ that is dense in $Q$.
We assert that $f(Q_{1})\subseteq f(Q)$ is a countable subset that
is dense in $f(Q$). Clearly $f(Q_{1})$ is countable. Let $y\in f(Q)$,
then there exists $x\in Q$ such that $y=f(x)$. Choose a sequence
$(x_{n})$ in $Q_{1}$ such that $x_{n}\rightarrow x$. By continuity
of $f$, we have $f(x_{n})\rightarrow f(x)=y$. Observe that $f(x_{n})\in f(Q_{1})$.
This shows that $f(Q_{1})$ is dense in $f(Q)$.
A: I'll provide an alternative proof.  Let $\emptyset \neq V \subseteq f(Q)$ be open in $f(Q)$.  Then for some $U \subseteq Y$ open, $V = U \cap f(Q)$.
Thus, $f^{-1}(U)$ is open in $X$ because $f$ is continuous, so $f^{-1}(U) \cap Q = f^{-1}(V) \cap Q$ is open in $Q$
Let $A \subseteq Q$ be countable and dense in $Q$.  Then $A \cap f^{-1}(V) \neq \emptyset$ because $A$ is dense in $Q$.  It follows, then, that $f(A) \cap V \neq \emptyset$.  Since $V$ was an arbitrary nonempty open subset of $f(Q)$, this means that $f(A)$, which is countable, is dense in $f(Q)$.
