TITLE: Hausdorff open image of a Polish space
QUESTION [9 upvotes]: Let $f\colon X\to Y$ a continuous open and surjective function, where   $X$ is Polish. 
It is known that $Y$ is Polish if: $f$ is closed or $Y$ is metric.
Suppose that we know that $Y$ is Hausdorff, does it implies that $Y$ is Polish? or Regular?
I think no, but i have not been able to find a counterexample.

REPLY [4 votes]: As a counterexample one can consider the projective space $P\mathbb R^\omega$ of the countable product of lines. This is a quotient space of the Polish space $\mathbb R^\omega_\circ=\mathbb R^\omega\setminus\{0\}^\omega$ by the equivalence relation $x\sim y$ iff $\mathbb Rx=\mathbb Ry$.
It is easy to see that the quotient map $q:\mathbb R^\omega_\circ\to P\mathbb R^\omega$ is open and the space $P\mathbb R^\omega$ is Hausdorff but not Urysohn and hence is not metrizable and not Polish.
Moreover, the space $P\mathbb R^\omega$ is superconnected in the sense that for any nonempty open sets $U_1,\dots,U_n$ in $P\mathbb R^\omega$ the intersection of thier closures $\bar U_1\cap\dots\cap\bar U_n$ is not empty.