The set $\{ g \in G | gK_1 \cap K_2 = \emptyset \}$ is open I am struggling in trying to prove the following statement:

Let $G$ be a topological group, acting continuously on a topological Hausdorff space $X$. Let $K_1, K_2 \subseteq X$ be two compact subsets of $X$. Then the set 
  $$H=\{ g \in G \ | \ gK_1 \cap K_2 = \emptyset \}$$ is an open subset of $G$.

I am sure this is a standard exercize in general topology, however I don't find any proof on the web.
My try: I fix $g_0 \in H$ and try to prove that there is an open neighbourhood of $g_0$ contained in $H$.
I know that $g_0K_1$ and $K_2$ are disjoint compact subsets of $X$. Since $X$ is Hausdorff, there exist two disjoint open sets $U$ and $V$ separating them, in other words
$$g_0K_1 \subseteq U \qquad \mathrm{and } \qquad K_2 \subseteq V \qquad \mathrm{and } \qquad U \cap V = \emptyset$$
Now I consider the set
$$W=\{ g \in G \ | \ gK_1 \subseteq U \}$$
Clearly $g_0 \in W$. Moreover, for all $g \in W$
$$gK_1 \cap K_2 \subseteq U \cap V = \emptyset$$
i.e. $W \subseteq H$. It remains to show that $W$ is open: however this looks as hard to prove as proving the original statement. And here I got stuck.
 A: You are almost there. Focus on the $g_0K_1\subseteq U$ condition. The idea is to enlarge $\{g_0\}$ to some open neighbourhood $V$ such that $VK_1\subseteq U$:

Lemma. Let $X$ be a $G$-space, $K\subseteq X$ its compact subspace and $gK\subseteq U$ for some $g\in G$ and open subset $U\subseteq X$. Then there exists an open neighbourhood $V\subseteq G$ of $g$ such that $VK\subseteq U$.

Proof. Consider the action $A:G\times X\to X$ of $G$ on $X$ and consider $A^{-1}(U)$. Take $x\in K$. Since $A^{-1}(U)$ is open then by the product topology we can choose a cube $V_x\times U_x\subseteq A^{-1}(U)$ such that $(g,x)\in V_x\times U_x$ and both $V_x$ and $U_x$ are open in their respective spaces. Note that $\{U_x\}_{x\in K}$ is a covering of $K$ and since it is compact then we can choose a finite number of points $x_1,\ldots, x_n\in K$ such that $\{U_{x_i}\}_{i=1}^n$ covers $K$. Now put $V:=V_{x_1}\cap\cdots\cap V_{x_n}$. It follows that $V$ is an open neighbourhood of $g$ and $VK\subseteq U$ because $VU_{x_i}\subseteq U$ for any $i=1,\ldots,n$. $\Box$
