I have been asked to show that if $X$ is a dense $G_\delta$ subset of $\mathbb{R}$ such that $\mathbb{R}\setminus X$ is also dense in $\mathbb{R}$, then $X$ is homeomorphic to $\mathcal{N}=\mathbb{N}^\mathbb{N}$ with the product topology. I was able to prove this using the fact that $\mathcal{N}$ is the only Polish zero-dimensional space in which every compact subset has empty interior, up to homeomorphism. However, I also want to show that this does not hold in $\mathbb{R}^2$, so I've considered the space $X=\mathbb{R}\times(\mathbb{R}\setminus\mathbb{Q})$, which does not look homeomorphic to $\mathcal{N}$. However, I don't really know if this is the case, but if it is, it is worth noting that $X$ is Polish and that every compact subset of $X$ has empty interior. I've been trying to show that $X$ is not zero-dimensional, but I haven't had any success.
Any help will be highly appreciated.
Your example works: consider the connected components.
For another, more extreme example, consider the complement of $\mathbf Q^2$.