Continuous images of open sets are Borel? Consider a Polish space $(X,d)$ and any metric space $(Y,e)$. If we have a continuous surjection $f:X\to Y$ then is the image $f(U)$ of any open subset $U\subset X$ a Borel set in $Y$? 
I know that this is true if we allow $X$ to be compact, since every open subset of a metric space is $F_{\sigma}$ and closed subsets of a compact metric space are compact, and moreover continuous images of compact sets are compact. So in this case, the image of an open set would be $F_{\sigma}$ in $Y$. 
If this is not the case for the given $f$, would it do any difference if we allowed $Y$ to be in addition Polish? Or compact? Or should $f$ have more properties? How about $X$? Or is this all just a triviality that I have overlooked?
Thanks for all the input in advance.
 A: I see that Michael Greinecker has preempted the idea on which I was working; I post this only because it contains a little more detail about the retopologizing, since I didn’t know the general lemma that Michael quotes and had to make my own way at that point.
Let $\mathscr{N}=\omega^\omega$ with the product topology $\tau$. Let $d$ be any complete metric on $\omega^\omega$ that generates $\tau$ and is bounded by $1$. Let $F$ be a closed subset of $\mathscr{N}$, and define a new metric $\rho$ on $\omega^\omega$ as follows:
$$\rho(x,y)=\begin{cases}
d(x,y),&\text{if }x,y\in F\text{ or }x,y\in\mathscr{N}\setminus F\\
2,&\text{otherwise}\;.
\end{cases}$$
It’s easy to check that $\rho$ is a metric. Let $\tau_\rho$ be the topology on $\omega^\omega$ generated by $\rho$, and let $X$ be $\omega^\omega$ with the topology $\tau_\rho$. Clearly $F$ is clopen in $X$. Any $\rho$-Cauchy sequence must be eventually in $F$ or in $X\setminus F$ and must therefore be $d$-Cauchy, so $\langle X,\rho\rangle$ is complete. Moreover, $\tau_\rho$ is just the join of $\tau$ and $\{\varnothing,F,\mathscr{N}\setminus F,\mathscr{N}\}$, so it’s second countable, and $X$ is therefore a Polish space in which $F$ is a clopen set.
Now let $A\subseteq\mathscr{N}$ be analytic but not Borel. Because $A$ is analytic, there is a closed $H\subseteq\mathscr{N}\times\mathscr{N}$ such that $A=\pi_0[H]$, where $\pi_0:\mathscr{N}\times\mathscr{N}\to\mathscr{N}:\langle x,y\rangle\mapsto x$. Let $h:\mathscr{N}\to\mathscr{N}\times\mathscr{N}$ be a surjective homeomorphism, and let $F=h^{-1}[H]$. Apply the construction of the first paragraph to this $F$. Then $f\triangleq\pi_0\circ h:X\to\mathscr{N}$ is continuous, and $f[F]=A$, so $A$ is a non-Borel image of the clopen set $F$ in the Polish space $X$.
A: The answer is no. Endow both $\mathbb{R}^2$ and $\mathbb{R}$ with the usual Borel $\sigma$-algebra and let $\pi_1:\mathbb{R}^2\to\mathbb{R}$ be the projection onto the first coordinate. Now it is a well known fact that there exists Borel sets $B\subseteq\mathbb{R}^2$ such that $\pi_1(B)$ is not Borel. Now $B$ will be in general not open, but we can make it so. We just apply the following theorem (Lemma 4.58 in Alipranits & Border 2006):
Let $\mathcal{C}$ be a countable family of Borel subsets of a Polish space
$(X, \tau)$. Then there is a Polish topology $\tau'\supseteq\tau$ on $X$ with the same Borel $\sigma$-algebra for which each set in $\mathcal{C}$ is both open and closed.
In particular, we can make $B$ open. Then $B$ is open and $\pi_1$ is a continuous  surjection, but $\pi_1(B)$ not Borel. Since one can take $B$ in the original space to be bounded, the example works even when $Y$ is compact.
A: On the other hand if $Y$ is Polish and $f$ is a countable to one Borel map, then the image of every Borel set is Borel.
