Given a group action $G\curvearrowright X$.
Consider the orbit space: $\pi:X\to X/G$
Do continuous group actions correspond to open projections, i.e.:
$$l_g\in\mathcal{C}(X)\quad(g\in G)\iff\pi(U)\in\mathcal{T}_{X/G}\quad(U\in\mathcal{T}_X)$$
(Note that this is a slightly different more appropriate version of continuous group actions.)
Certainly, continuous group actions give rise to open projections since:
$$\pi^{-1}(\pi(U))=\bigcup_{u\in U}Gu=GU=\bigcup_{g\in G}gU=\bigcup_{g\in G}l_g(U)\in\mathcal{T}$$
Surely, the converse may hold as the following example suggests:
$$l_{k\in\mathbb{Z}}:=\mathrm{id}+k\chi_\mathbb{Z}:\quad \pi^{-1}(\pi(-\varepsilon,\varepsilon))=(-\varepsilon,\varepsilon)\cup\mathbb{Z}\notin\mathcal{T}_\mathbb{R}$$
Besides, the analog assertion about closed projections is wrong:
$$l_{q\in\mathbb{Q}}(x):=x+q:\quad\pi^{-1}(\pi(\{0\}^\complement))=\mathbb{Q}^\complement\notin\mathcal{T}_\mathbb{R}$$
(That is arbitrary unions of closed sets are not closed in general.)
Still how to prove that the converse holds true, too?
Reference: This is a follow-up to: Group Actions: Discontinuity
Best Answer
This is false in general, take for instance, $G\cong {\mathbb Z}$ acting on the circle via irrational rotations. Then the projection of each point to $Q=S^1/G$ is not only not closed, it is actually a dense point in $Q$.
Edit: I misread the original question as I thought you were asking about the quotient map being closed. The answer to you question is that open projection $\pi$ does not imply that the action is continuous. As an example, consider the transitive action of the group $G$ of real numbers on itself via left translation, $G\times X\to X$, $X=G$. Now, change the topology on $X$ to make this action discontinuous (say, pick a bijection of $f: X\to [0,1]$ and equip $X$ with the preimage of the standard topology on $[0,1]$). The resulting action is no longer continuous, but the map $\pi: X\to X/G=\{p\}$ is open.