Does a continuous function embed a separable space into a separable closed subspace?

Question

Suppose that $X$ is a topological space and $Y$ is a normed vector space, and $f:X\rightarrow Y$ is continuous.

In general we know that if $X$ is separable, then the image $f[X]$ will be separable in $Y$. Does it then follow that there exists a closed subspace $Y_0$ of $Y$ which is separable and contains the $f[X]$?

(Basically I am asking this question with an eye towards the Pettis measurability criterion from Bochner integration.)

(Edited to fix a typo)

Answer 1

There is a closed (separable) subset of $Y$ that contains $f[X]$ namely $\overline{f[X]}$.

The closure of a separable subspace is still separable.