Let $p$ be the quotient map. Then $p^{-1}([Gx])=G\cdot x$ , compact as the continuous image of the compact space $G$ . Suppose now, that $A\subset X$ is closed. We show that $G\cdot A$ is closed. This would imply that $p(A)=[GA]$ is closed, therefore, $p$ is a closed quotient map, and we can use the previous two exercises to prove what we are asked to (instead, as suggested in the hint, one could try to use Exercise 13(b) of §26). Let $z\notin G\cdot A$ . For every $g\in G$ , $g\cdot z\notin A$ . Since $\alpha$ is continuous and $X-A$ is open, $\alpha^{-1}(X-A)$ is open, and there is some basis neighborhood $U_{g}\times V_{g}$ of $g\times z$ such that $\alpha(U_{g}\times V_{g})$ does not intersect $A$ . The union of all $U_{g}$ covers $G$ . Since $G$ is compact, take a finite subcovering, the corresponding finite intersection $V$ of $V_{g}$ is a neighborhood of $z$ such that $\alpha(G\times V)$ does not intersect $A$ . Therefore, $V$ does not intersect $G\cdot A$ .