Homeomorphic and space Polish Let $X \subseteq  \Re$ be $G_{\delta}$ and such that $X$, $\Re \setminus X$ are dense in $ \Re$. Then $X$ is homeomeorphic to $\mathcal{N}$. Also when replacing $ \Re$ by a zero-dimensional nonempty Polish space. Where $\Re$ are the reals and $\mathcal{N}$ is a space Baire.
As $\Re$ is a Polish space and $ X $ is $ G_{\delta} $ then $ X $ is a Polish subspace of $\Re$, if $X$ is closed then $X$ is homeomeorphic to $\mathcal{N}$? 
¿ As  ensure that $X$ is homeomeorphic to $\mathcal{N}$? 
 A: A separable completely metrisable $X$ is homeomorphic to $\mathcal{N}$ iff $X$ is zero-dimensional and nowhere compact (see references in this article). 
This immediately implies the statements for the reals and for zero-dimensional spaces: $X$ is a $G_\delta$ in a Polish space so Polish (completely metrisable) in both cases.
A dense subset $X$ of the reals such that its complement is also dense, must be zero-dimensional. To see this, every point $x \in X$ has neighbourhood of the form $(a,b) \cap X$, and as $\mathbb{R}\setminus X$ is dense, there are $c,d \notin X$ such that $c,d \in (a,b)$ and $x \in (c,d)$, and then $(c,d) \cap X$ is a clopen neighbourhood of $x$ inside $(a,b) \cap X$. For zero-dimensional (Polish) spaces, $X$ will be automatically zero-dimensional as a subspace of a zero-dimensional space.
Being nowhere compact means there for all compact set $K$ we have that $\operatorname{int}(K) = \emptyset$. It suffices to show that all non-empty sets of the form $O \cap X$ have non-compact closure in $X$, where $O$ is open in the surrounding space $S$ (either reals or a zero-dimensional Polish space). Pick $q \in O \cap (S \setminus X)$, by denseness. Then as $X$ is dense, we pick a sequence $x_n \in X \cap O$ such that $x_n \rightarrow q$. If $O \cap X$ would have a compact closure (in $X$!), the sequence has a convergent subsequence that converges to some $p \in X$. But unicity of limits says $p = q$ and $q \notin X$, so this cannot happen. So a dense subset with dense complement is indeed nowhere compact.  
A: If $X$ is closed and dense then $X=\frak R$; in which case $\mathfrak R\setminus X$ cannot possibly be dense.
