Let ${\mathbb{N}^\mathbb{N}}^*$ denote the infinite sequences in Baire space that do not have a periodic tail, which is a $G_\delta$ subset so it is a Polish space. Consider the equivalence relation on ${\mathbb{N}^\mathbb{N}}^*$ where two sequences $a$ and $b$ are equivalent if and only if they share the same tail (they may have different lengths of initial segments) which is a Borel relation.
When I try to find a Borel dense co-dense subset $A$ of ${\mathbb{N}^\mathbb{N}}^*$, I feel like I need to put a restriction on infinitely many indices and this makes some equivalence classes to be contained in either $A$ or $A^c$. My question is, is there a Borel dense co-dense subset $A$ in ${\mathbb{N}^\mathbb{N}}^*$ such that for each equivalence class, some elements of the class are in $A$ and some are in $A^c$?
Edit: I could ask a similar question on $[0,1]$. Is there a Borel dense co-dense subset $A$ such that both $A$ and $A^c$ contain some points in each equivalence class of the Vitali equivalence relation.
Here's an example for $[0,1]$ (per your edit) that lifts without serious difficulty to the original question.
We start with $A_0=[0,{1\over 2}]$. Clearly both $A_0$ and $[0,1]\setminus A_0$ meet every Vitali class. So now we just need to modify $A_0$ to get "bi-density" without damaging this.
To do this, fix any countable dense $C\subseteq[0,1]$ and let $$A=(A_0\setminus\{{c\over 2}: c\in C\})\cup \{{1+c\over 2}: c\in C\}.$$ Since $c\over 2$ is Vitali-equivalent to ${1+c\over 2}$ for any $c$, the new set $A$ also meets each Vitali class and has complement meeting each Vitali class. And since $C$ is countable (so a fortiori $F_\sigma$), the set $A$ is Borel.