Proposition: Let $R$ be an equivalence relation on the space $X$, if the equivalence relation $R$ is minimal, then the only subsets of $X$ which are both closed, open and $R$-invariant are $X$ and the empty set.if $R=R_\phi$, where $\phi$ is a free action of the group $G $ on $X$, then the converse also holds.
Def'n Let R be an equivalence relation on the space X. $ A\subset X $ is $R$-invariant if whenever $x\in A $, it follows that $[x]_R\subset A$
Def'n Let R be an equivalence relation on the space X. we say that R is minimal if, for every $x\in X $, its R-equivalence class, $[x]_R$ is dense in X.
Def'n Let $\phi $ be an action of the group G on the space X. we deifne the orbit relation of $\phi $ to be $R_\phi =\{(x,\phi^{g}(x)) | x\in X, g\in G \} $
Let $R$ be an equivalence relation on the space $X$ where the equivalence relation $R$ is minimal but $\exists A \subsetneq X | A \neq \emptyset$ which is closed and $R$-invariant. Since A is closed $ \overline A=A$ but $\overline {[x]_R}=X$ but this a contradiction as $ \overline {[x]_R}\subset \overline A $ hence the result follows.
I have no idea how to prove the converse of the proposition when $R=R_\phi$ any help much appreciated