Every second-countable $T_0$ space is the quotient of a separable metric space and this essentially follows from the proof of the $T_1$ case given by Paul Strong in [*Quotient and pseudo-open images of separable metric spaces*](http://www.ams.org/journals/proc/1972-033-02/S0002-9939-1972-0300253-7/home.html) [Proc. Amer. Math. Soc. 33 (1972), 582-586].

A continuous map $f:X \to Y$ is *sequence covering* if for every convergent sequence $y_n \to y$ in $Y$ there is a convergent sequence $x_n \to x$ in $X$ such that $y_n = f(x_n)$ for all $n$ and $y = f(x)$. (Since the spaces under consideration are not necessarily Hausdorff, when I say "convergent sequence" I always mean a convergent sequence together with a choice of limit.)

**Fact 1.** *If $f:X \to Y$ is a sequence covering continuous map and $Y$ is sequential then $f:X \to Y$ is a quotient map.*

Let $\omega+1 = \{0,1,\dots,\omega\}$ be the one-point compactification of $\omega$. The space $Y^{\omega+1}$ with the compact-open topology consists of all convergent sequences in $Y$ (along with a choice of limit). Since $\omega+1$ is compact Hausdorff, the evaluation map $e:(\omega+1)\times Y^{\omega+1}\to Y$ is continuous. Moreover, this map it is obviously sequence covering. Even better:

**Fact 2.** *If $g:X \to Y^{\omega+1}$ is a continuous surjection then the continuous map $f:(\omega+1)\times X \to Y$ defined by $f(n,x) = e(n,g(x))$ is sequence covering.*

Combining these facts, we see that if $Y^{\omega+1}$ is sequential and the continuous image of a separable metric space $X$ then $Y$ is the quotient of the separable metric space $(\omega+1)\times X$.

Now if $Y$ is $T_0$ and second-countable then so is the space $Y^{\omega+1}$. Therefore, it suffices to prove:

**Fact 3.** Every second-countable $T_0$ space $Y$ is the continuous image of a separable metric space.

Tho see this, fix a countable base $(U_n)_{n\lt\omega}$ for $Y$. Let $X \subseteq \omega^\omega$ consist of all $x:\omega\to\omega$ such that $(U_{x(n)})_{n \lt\omega}$ enumerates a neighborhood basis for some point of $Y$. Since $Y$ is $T_0$, a neighborhood basis for a point uniquely determines that point. Therefore, we obtain a natural surjection $f:X \to Y$, which is easily seen to be continuous.

Since a second-countable space is always sequential, we can combine the three facts above to conclude that every second-countable $T_0$ space is a quotient of a separable metric space.