Every 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. 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.