Continuous open metrizable image of a Polish space is Polish

If $X$ is Polish, it embeds as a $G_\delta$ in the Hilbert cube $[0,1]^\omega$. But the open map doesn't help us there, because it concerns sets around $X$. So we need an internal characterisation in terms of open sets preferably.

Well, a metrisable space is completely metrisable iff it's Čech-complete and that has the following characterisation:

$X$ has a family of open covers $\mathcal{U}_n, n \in \omega$ such that every family of closed sets $\mathcal{F}$ that has the finite intersection property and such that $$\forall n \in \omega: \exists F \in \mathcal{F}: \exists U \in \mathcal{U}_n: F \subseteq U$$ then $\bigcap \mathcal{F} \neq \emptyset$.

This compact-like property is equivalent to Čech-completeness in Tikhonov spaces and so equivalent to complete metrisability if $X$ is already known to be metrisable. It was a tempting idea to just to take the images under $g$ of these open covers for the completely metrisable $X$ to get open covers of $Y$. But $F \subseteq g[U]$ does not always imply $g^{-1}[F] \subseteq U$, so that idea didn't work either (I couldn't show the crucial property for the covers in $Y$ from that property in $X$, but maybe someone else sees an easy way to "fix" this using paracompactness or suitable refinements.

But then I found in Engelking (to find the example of non-Čech-complete space which is the open image of a Čech-complete space) the following theorem

[Pasynkov 1967]: If $f:X \to Y$ is open continuous onto $Y$ and $X$ is locally Čech-complete and $Y$ is paracompact, then $Y$ is Čech-complete.

From this your desired result follows quite easily. But further on in Exercise 5.5.8 in Engelking was exactly what we need, due to Michael and earlier Hausdorff and Sierpiński (1934 resp 1930 for more restrictive classes):

Michael's 1959 version: If $f:X \to Y$ is open continuous and onto, and $X$ is completely metrisable and $Y$ is paracompact (Hausdorff) then $Y$ is completely metrisable.

The paper is: A theorem on semi-continuous set valued functions, Duke Math. J. 26 (1959), 647-651 (part of a nice series of papers on selection theorems, real classics, BTW)

This may not be the answer you want, since I am only providing further reading, but the following is a result of Hausdorff (1934).

Theorem A. If $X$ is completely metrizable, and there is a continuous open map $f:X\to Y$ from $X$ onto a metrizable space $Y$, then $Y$ is completely metrizable.

I now quote from A Note on Completely Metrizable Spaces by E. Michael, where a different proof of this result is discussed.

The simplicity of this argument contrasts strikingly with Hausdorff's complicated and lengthy proof in [H2].

So you may want to check out the details in Michael's paper. In particular, the proof discussed there goes through the notion of a "complete open sieve". It is shown (in earlier works) that a metrizable space is completely metrizable if and only if it has a complete open sieve. So, in Theorem A, $X$ has a complete open sieve, which is mapped to a complete open sieve in $Y$ under $f$, hence $Y$ is completely metrizable. I did not look further into these details, but perhaps in the setting that $X$ is Polish it is easier to describe the complete open sieve in question.

As a final remark, a generalization of Theorem A to the case that $Y$ is paracompact is given as Exercise 5.5.8(d) in General Topology by Engelking.