The definition of a topological group $G$ acting on a topological space $X$ is there exists a continuous map from $G\times X \rightarrow X$ such that $e_G.x=x$ for all $x\in X$ and $(g_1.g_2)(x)=(g_1.(g_2.x))$.
It follows from the definition that if $G$ acts on $X$ then we have a group homomorphism from $G\rightarrow Homeo(X)$.
My question is: Is this true that if there is a group homomorphism $G\rightarrow Homeo(X)$ then $G$ acts on X.
Note: We just have a group homomorphism from $G$ to $Homeo(X)$, not a continuous map. I could prove that G acts on X on X as a set (i.e., we have map from $ G\times X \rightarrow X$ such that $e_G.x=x$ for all $x\in X$ and $(g_1.g_2)(x)=(g_1.(g_2.x))$) . But I dont see why that action will be continuous.
$\text{Homeo}(X)$ has a natural topology - the compact-open topology. What you want to demand is that the map $G \to \text{Homeo}(X)$ is continuous. (This, at the very least, works for reasonable groups, like locally compact ones. I don't know at what generality a continuous action $G \curvearrowright X$ is the same as a continuous homomorphism $G \to \text{Homeo}(X).$)
As an example of a discontinuous group action, pick a discontinuous group homomorphism $\Bbb R \to \Bbb R$, considering $\Bbb R$ as the subgroup of $\text{Homeo}(\Bbb R)$ with $c \in \Bbb R$ defining the homeomorphism $c(x) = x+c$. There are lots of discontinuous group homomorphisms here. You can verify (just look at what happens to a single point!) that such a thing defines a discontinuous group action of $\Bbb R$ on $\Bbb R$.