Let $F$ be a topological space such that $G := Homeo(F)$ is a topological group in the compact-open topology. Let $\phi : G \times F \to F$ sending $(g,x)$ to $g(x)$ be the map for the action of $G$ on $F$. I'm wondering if this action is continuous. I can't seem to prove it nor can I find counterxamples, but here is what I have so far.
Let $U$ be open in $F$. I've shown that $\phi^{-1}(U) = \cup_{g \in G} \{g\}\times g^{-1}(U)$. Since $g$ is a homeomorphism, $g^{-1}(U)$ is open in $F$. I'm now struggling to use the compact-open topology to show that the union over $g$ is open, if this is even true.
Or do I need to assume more on $F$ or $G$ to make the group action continuous?
EDIT: In the comments we have that $F$ being locally compact as well is sufficient for $G$ to act continuously. But is it necessary?