How could I prove this proposition? Please help me.
Orbit space of a continuous group action. Suppose a right action of a topological group $G$ on a topological space $S$ is continuous; this simply means that the map $S\times G\to S$ describing the action in continuous. Define two points $x, y$ of $S$ to be equivalent if they are in the same orbit; i.e., $\exists g\in G\, \text{s.t.}\, y = xg$. Let $S/G$ be the quotient space; it is called the orbit space of the action. Prove that the projection map $\pi: S\to S/G$ is an open map. (This problem generalizes Proposition $7.14$, in which $G = R^\times = \mathbb R - \lbrace 0\rbrace$ and $S = \mathbb R^{n + 1} - \lbrace 0\rbrace$. Because $\mathbb R^\times$ is commutative, a left $\mathbb R^\times$-action becomes a right $\mathbb R^\times$-action if scalar multiplication is written on the right.)
Suppose $O \subseteq S$ is open. As $X{/}G$ has the quotient topology w.r.t. $\pi$, $\pi[O]$ will be open iff $\pi^{-1}[\pi[O]]$ (the saturation of $O$ under the equivalence relation) is open in $S$.
Now note that
$$\pi^{-1}[\pi[O]] = \{x \in S: \exists x' \in O: \exists g \in G: gx'=x\}$$
which equals $$\bigcup_{g \in G} Og= \bigcup T_g[O]$$
where by $Og$ I mean all points $\{x \cdot g, x \in O\}$ where $\cdot$ denotes the group action, and $T_g: S \to S, T_g(x)=x \cdot g$ is a homeomorphism (continuous as the action is, and with (also continuous) inverse $T_{g^{-1}}$). So $\pi^{-1}[\pi[O]]$ is a union of open sets (images of $O$ under the homeomorphisms $T_g$, which are open maps) hence open, and so $\pi$ is an open map too.