Every [analytic set](http://en.wikipedia.org/wiki/Analytic_set) ($\Sigma^1_1$ set) of reals is the projection of a Borel subset of $\mathbb{R}\times\mathbb{R}$, and the projection map $p(x,y)\mapsto x$ is an open map. So the standard examples of non-Borel $\Sigma^1_1$ sets are also examples where Borel sets are not preserved by an open map $\mathbb{R}^2\to\mathbb{R}$.

But you asked for an open map $\mathbb{R}^n\to\mathbb{R}^n$, with the same domain and codomain, and the reasoning above concerned only an open map $\mathbb{R}^2\to\mathbb{R}$. Here is one way to fix the issue and make an open map $\mathbb{R}^3\to\mathbb{R}^3$ having the image of a Borel set being non-Borel. Let $h:\mathbb{R}\to\mathbb{R}^2$ be any function whose restriction to every open interval is onto. One can make such a function by using Cantor's interleaving digits trick, combined with the idea of [Conway's base 13 function](http://en.wikipedia.org/wiki/Conway_base_13_function). This function is an open map, since every nonempty open set maps onto the whole space. Now, define $f(x,y,z)=(x,z_0,z_1)$, where $h(z)=(z_0,z_1)$. It is easy to see that the function $f$ is an open map. Meanwhile, every analytic set $A$ has the form $x\in A\iff \exists y B(x,y)$, where $B\subset\mathbb{R}^2$ is a Borel set. Let $C=B\times\mathbb{R}$, which is Borel. Consider the image set $f[C]$, and note that $(x,0,0)\in f[C]$ if and only if there is some $y$ such that $(x,y)\in B$, since in this case we will find a $z$ with $h(z)=(0,0)$; hence, $(x,0,0)\in f[C]$ if and only if $x\in A$, and so the intersection of $f[C]$ with the $x$-axis is $A$, a non-Borel set. So $f[C]$ cannot be Borel if $A$ is not. So this is a case where we have an open map $f:\mathbb{R}^3\to\mathbb{R}^3$ taking a Borel set to a non-Borel set.