In Engelking's General Topology, Exercise $4.2.\text{D}.(\text{a})$, we are asked to show that a $T_0$ space is $1^{st}$ Countable iff it is the continuous image of a metrizable space under an open mapping. As a metric and pseudometric space differ only by the $T_0$ axiom, this made me wonder - can we remove the $T_0$ axiom and get a result for a pseudometrizable space? That is -
Is $X$ $1^{st}$ Countable iff it is the continuous image of a pseudometrizable space under an open mapping?
If you do part (b) of that exercise too, you see that if $X$ is first countable iff $X$ is the open continuous image of a metrisable space. This is theorem 4.3 in this paper by E. Michael.
And it's trivial that if $X$ is the open continuous image of a pseudometrisable space, then it is first countable. So yes, but not for the reason you thought maybe (it has nothing to do with the $T_0$ difference between pseudometrisable and metrisable spaces).