Suppose the space $X$ has a countable basis and $X$ is $T_{0}$. Must there exist a separable metrizable space $Y$ and a quotient map q:$Y \rightarrow X$?

(Some surrounding facts:

Every metrizable space is 2nd countable iff it's separable.

Every 2nd countable space is 1st countable and hence Frechet and hence sequential and hence the quotient of a locally compact metrizable space. ( However in the canonical proof, $Y$ is the disjoint union of a typically very large collection of convergent sequences [Franklin] and usually not separable, even if $X$ itself is a separable metric space).

If $X$ is $T_{0}$ and regular and 2nd countable then $X$ is metrizable (Urysohn metrization)).

For a non $T_{0}$ counterexample let $X$ have cardinality larger than the real numbers and employ the indiscrete topology.)

If the answer is `no' can a counterexample $X$ be $T_1$ or even $T_2$?

(Edit: the answer is `yes'. Solutions and relevant links appear below, plausibly hidden in the comments)