Consider a Polish space $X$, and let $G$ be a countable group which acts freely on $X$ (i.e. $gx\neq x$ for any $x\in X$ and $g\in G\setminus \{e\}$). Let $x\in X$ be a recurrent point for the G-action, which means there exist $g_n\neq e$ such that $g_n x\to x$ as $n\to\infty$. Let $y\in \overline{Gx}\setminus Gx$, where $Gx=\{gx:g\in G\}$ is the $G$-orbit of $x$. I'd like to show that $Gx$ and $Gy$ (which are distinct points of the quotient) cannot be separated (in $X/G$) in the $T_0$ sense.
If we take an open set $V\subset X/G$ such that $Gy \in V$, then $y\in \bigcup V$, which is open in $X$, and thus $gx\in \bigcup V$ for some $g\in G$ since $y$ is in the closure of the orbit. It follows that $Gx\in V$.
I'm having trouble proving that, if $V\subset X/G$ is open and $Gx\in V$, then $Gy\in V$. I've not yet used the fact that $x$ is a recurrent point, so I'm guessing that's somehow involved. It would suffice to show that $gy\in V$ for some $g\in G$, which seems reasonable since there are points of the form $gx$ arbitrarily close to $y$ in $X$, but I don't see how to go from a neighborhood of $x$ to one of $y$. The main issue seems to be showing that points of the $G$-orbit of $x$ cannot approach the boundary to $\bigcup V$. Equivalently, we would be showing that there exists a $G$-invariant set $W$ with $x\in W$ and $\overline{W}\subset \bigcup{V}$.
EDIT: I forgot to add, the existence of $y\in \overline{Gx}\setminus Gx$ is guaranteed by the fact that $x$ is recurrent, so this fact has actually been used at an onset. For, if $x$ is recurrent, then so is every point of $Gx$, and thus $\overline{Gx}$ is perfect, hence uncountable, since $X$ is Polish. But $Gx$ is countable (since $G$ is), and so $\overline{Gx}\setminus GX\neq \emptyset$ (it actually has the cardinality of the continuum, but that does not seem very relevant).
The claim is false and there are many natural examples. For instance, let $S$ be a compact surface of constant negative curvature which has no nontrivial closed geodesics of integer length, let $M$ be the unit tangent bundle of $S$ and let diffeomorphism $f: M\to M$ be the time 1 geodesic flow of $S$. Then $f$ is known to be ergodic (Grayson-Pugh-Shub, Annals of Math, 1994). The cyclic group $G$ generated by $f$ acts freely and ergodically on $M$, hence, almost every orbit is dense. On the other hand, there are points $y\in M$ whose orbits are not dense, for instance, one can take the unit vector tangent to a closed geodesic on $S$. Let $x\in M$ be a generic point with dense orbit such that $x\notin \overline{G y}$. Hence, $Gx$ accumulates to $x$ and to $y$ but $Gy$ does not accumulate to $x$.