Proof. (1) $\Rightarrow$ (2): Let $x \in X$, then $\ol{Gx}$ with the restricted action is a $G$-subflow of $X$. As $\emptyset \ne \ol{Gx}$ and $G \curvearrowright X$ admits no proper subflows, $\ol{Gx}= x$.

(2) $\Rightarrow$ (1): Let $G \curvearrowright Y$ be a $G$-subflow of $G \curvearrowright X$. For any $x \in Y$, $Y \supset \ol{Gx}= X$, so $Y = X$.

(2) $\Rightarrow$ (3): Let $\emptyset \ne U \subset X$ be open. Since $G \curvearrowright X$ is minimal, $Gx \cap U \ne \emptyset$ for all $x \in X$. Thus $X \subset G^{-1}U = GU$. By compactness of $X$, there exists $F \subset G$ finite such that $FU = X$.

Let $x \in X$ be arbitrary, then for each $g \in G$, $gx \in X = FU$. Thus there exists $h \in F$ with $gx \in hU$ and $h^{-1}gx \in U$. Hence $h^{-1}g \in N(x, U)$, and $g \in hN(x, U) \subset FN(x, U)$.$\square$