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' and Francois Dorais and Andrej Bauer provide two explicit solutions below and also point out relevant references.

The $T_{1}$ case was settled by Paul Strong. As shown below similar tactics settle the $T_{0}$ case. The question is relevant to topological domain theory. For example

``The similarity between our definitions and results and those of Schroder was first observed by Andrej Bauer, who proved that the sequential spaces with admissible representations are exactly the T0 (quotients of countably based) spaces...''.

From the paper `Topological and Limit-space Subcategories of Countably Based Equilogical Spaces.' by Menni and Simpson.

homepages.inf.ed.ac.uk/als/Research/Sources/subcats.pdf