Let G be a compact topological group and X be a topological space. The map $f: G\times X \to X$ is continous. Prove that for each open subset $U\subset X $ the set $V=\{x \in X\mid f(x,g)\in U, \forall g\in G\}$ is open in X.
First we know since $f$ is continous, that $f^{-1}(U)$ is open in $X \times G $. Say $Y \times H=f^{-1}(U)$ since in the product topology a set is open when $Y$ and $H$ are open in there topologies. So $Y=\{x \in X\mid \exists g\in H, f(x,g)\in U\}$ is open in S.
At that point I am not sure how to continue.
First $f^{-1}(U)$ does not have to be of the form $Y\times H$. Think of the open ball in $\Bbb R^2$. You should take a look at definition of the product topology. (The products of open sets form a base of the topology.)
Assume $G\times \{x \}$ is contained in $f^{-1}(U)$, then there is also a open neighborhood of $G\times \{x \}$ in $f^{-1}(U)$. I.e. $G\times \{x \} \subset \bigcup V_i\times U_i \subset f^{-1}(U)$ where $V_i\subset G$, $\{x\}\subset U_i$. But since $G$ is compact we know that we can assume the index set to be finite. Thus $G\times (\bigcap U_i ) \subset f^{-1}(U)$ and $(\bigcap U_i )$ is an open neighborhood of $x$ contained in $V$, thus $V$ is open.