Suppose that $X$ is a topological space, and $T$ is a topological group which continuously acts on $X$ on the right. We call the pair $(X,T)$ a (right) transformation group. Now suppose that $\equiv$ is an “invariant” equivalence relation on $X$ in the sense that $\equiv$ is an arbitrary equivalence relation such that $x.t\equiv y.t$ whenever $x\equiv y$ for all $t\in T$ and for all $x,y\in X$.
It is easy to see that $(X/\equiv, T)$ is a transformation group with the action of $T$ defined by $[x].t=[x.t]$ where we have assumed $T$ to be given the discrete topology.
However, why is $(X/\equiv,T)$ a transformation group with the original topology of $T$ if the equivalence relation is open or if $X$ is compact and Hausdorff and $\equiv$ is a closed subset of $X\times X$? Even the theorems needed to see why this is true suffices for me to try proving it.
I will use the notation that a subset of $X/\equiv$ is denoted by $[V]$ where the set $V$ is $\{x\in X\mid [x]\in[V]\}.$ I found this more convenient than referring to sets of equivalence classes. Also, to me "neighborhood of $x$" means "set whose interior contains $x$" - a neighborhood doesn't have to be open.
We need to show that for each $(x,t)\in X\times T$ and each neighborhood $[U]$ of $[x]\cdot t,$ there is a neighborhood $[V]$ of $[x]$ and a neighborhood $S$ of $t$ such that $[V]\cdot S\subseteq [U].$
Suppose the equivalence relation is open. Using continuity of the action map there are neighborhoods $V'$ of $x$ and $S$ of $t$ such that $V'\cdot S\subseteq U.$ Set $V=\bigcup_{v\in V'}\{x\mid x\equiv v\}.$ Then $V$ is open so $[V]$ is open. And $V\cdot S\subseteq U$ as required.
Suppose the equivalence relation is closed and $X$ is compact Hausdorff. The set $U\cdot t^{-1}$ is open, because $y\mapsto y\cdot t^{-1}$ is a homeomorphism. And $U\cdot t^{-1}$ is a union of $\equiv$ equivalence classes. So $[U\cdot t^{-1}]$ is open. $X/\equiv$ is compact Hausdorff - see Question about quotient of a compact Hausdorff space for example. Since compact Hausdorff spaces are regular, there is a closed neighborhood $[K]$ of $[x]$ with $[K]\subseteq [U]\cdot t^{-1} = [U\cdot t^{-1}].$ For each $k\in K$ there are open sets $K_k\subseteq X$ and $T_k\subseteq T$ with $1\in T_k$ and $K_k\cdot T_k\subseteq U\cdot t^{-1},$ by continuity of the action map. The set $K$ is a closed subset of $X,$ so it is compact. There is therefore a finite set $K'$ such that $\bigcup_{k\in K'} K_k$ covers $K.$ The set $T'=\bigcap_{k\in K}T_k$ is an open neighborhood of $1\in T$ and satisfies $K\cdot T'\subseteq U\cdot t^{-1},$ i.e. $[K]\cdot (T't)\subseteq [U].$