In their 2006 paper "Turbulence, amalgamation, and generic automorphisms of homogeneous structures" Kechris and Rosendal (see here for the arXiv version of the paper) state the following proposition without proof:
Proposition 1.4. Let $G$ be a closed subgroup of $S_\infty$ and suppose $G$ acts continuously on the Polish space $X$. Then the following are equivalent for any $x \in X$:
- the orbit $G \cdot x$ is dense $G_\delta$;
- $G \cdot x$ is dense and turbulent.
They say that this can easily be proved directly but I am quite stuck. So any hints on how to prove this are appreciated.
Note that we call a point $x$ in $X$ turbulent if for every open neighbourhood $U$ of $x$ and every symmetric open neighbourhood $V$ of the identity of $G$ the local orbit $\mathcal{O}(x,U,V)$ is somewhere dense, i.e. $\mathrm{Int}(\mathrm{Cl}(\mathcal{O}(x,U,V))) \neq \emptyset$. Here, $\mathcal{O}(x,U,V) = \{ y \in X : \exists g_0,\dots,g_k \in V \forall i < k (g_i g_{i-1}) \dots g_0 \cdot x \in U \text{ and } g_k g_{k-1} \dots g_0 \cdot x = y \}$. This notion is $G$-invariant, i.e., we can speak about tubulent orbits.
In the paper linked above you can find several characterisations of turbulence in Proposition 3.2. One useful characterisation might be that $G \cdot x$ is turbulent if and only if it is $G_\delta$. However, I still don't succeed in proceeding from here.