No.  Projection from $\mathbb R^2$ onto $\mathbb R$ is open, but the image of a $G_\delta$ set can be any analytic set.  So there are non-Borels among them.  See http://en.wikipedia.org/wiki/Analytic_set  

**edit**  Not a counterexample from a space to itself as required.  

So a counterexample has to be an open map $\mathbb R^n$ to itself, but *not* at-most-countable-to-one, since those maps do preserve Borel, as I recall.