$G$ finite subgroup of $\operatorname{Diff}(M)$. The open $G$-stable subsets of $M$ form a base to the topology of $M$. Let $G$ be a subgroup of $\operatorname{Diff}(M)$ (the diffeomorphism group of a manifold $M$). A subset $S\subset M$ is said to be stable if it is connected and if we have either $gS=S$ or $gS\cap S=\varnothing$. Actually, I have two questions.
Question 1. According to this book, the $G$-stable subsets of $M$ are exactly the components of $G$-invariant subsets of $M$. But, I cannot see how that is true, since if $gS\cap S=\varnothing$, for a $g\in G$, how can $S$ be $G$-invariant?
Question 2. The book also says that the open $G$-stable subsets form a base for the topology of $M$, since for every $x\in M$ there is a arbitraly small open $G$-subset $S$, s.t. $G_x=G_S$; where $G_S=\{g\in G: gS=S\}$. How can I find such a $S$?
 A: Question 1
As said in the commentaries, a component of a $G$-invariant set does not neet to be $G$-invariant. It follows from the fact that image of a connected subset of a Manifold under continuous function is connected. Thus, if $S$ is a component of a $G$-invariant subset and $gS\cap S\neq\varnothing$, then $gS\subset S$. Applying the same argument to $g^{-1}$, one has that $gS=S$, if $gS\cap S\neq\varnothing$. Thus, the components of invariant subsets are $G$-stable. On the other hand, it is clear that $S$ is the component of $G(S)=\{gx:g\in G,\, x\in S\}$; which is $G$-invariant.
Question 2
If $G<\operatorname{Diff}(M)$, then we can choose a $G$-invariat metric on $M$. Also, for each $x\in M$ the exponential map associated to this metric gives a diffeomorphism $\exp_x:B(0,\epsilon)\subset T_xM\to W$, in which $W$ is a nhood of $x$ in $M$. As the metric is $G$-invariant, for each $g\in G_x$,  $(dg)_x$ is an orthogonal transformation of $T_xM$ and $exp_x\circ(dg)_x=g\circ \exp_x$. So, $(dg)_xB(0,\epsilon)=B(0,\epsilon)$, which means that $gW=W$. On the other hand, if $g\notin G_x$, then as $\exp_x\circ(dg)_x=g\circ \exp_{(gx)}$ and as $M$ is Hausdorff, we have (by shrinking) $W$ if necessary, that $gW\cap W=\varnothing$.
