The following is exercise 6 from chapter I of Mac Lane and Moerdijk's Sheaves in Geometry and Logic:
I'm currently having trouble with point $(b)$, probably because of my little ability with group actions and topological groups in general. We are asked to define a functor $r_G$ sending a set $X$ with an action $X\times G\rightarrow X$ to the set $r_G(X)$: the restriction of the action is now continuous. The trouble is: how can I define, for an equivariant map $f:X\rightarrow Y$, the corresponding $r_G(f)$? The only obvious choice is the restriction $f_{|r_G(X)}$, but I see no reason why if $x\in r_G(X)$ then $f(x)\in r_G(Y)$.
(the result is also mentioned in "category of G-sets" in nLab but without any details)
Woops, that was easy: obviously $I_x\subseteq I_{f(x)}$, and if $h\in I_{f(x)}$ then $h\in h I_x\subseteq I_{f(x)}$ ($I_{f(x)}$ is a subgroup of $G$), but $h\cdot-$ is an open map and so $h$ is contained in an open subset of $I_{f(x)}$.