This is Exercise I.7 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]".
Here $\mathbf{B}G$ is the category of all continuous $G$-sets, where $G$ is a topological group.
The Question:
Show that the forgetful functor $U:\mathbf{B}G\to\mathbf{Sets}$ need not preserve infinite limits.
Thoughts:
I have considered the contrapositive: that if some $U':\mathbf{B}G\to\mathbf{Sets}$ preserves infinite limits, then it is not a forgetful functor. I don't think this is the best way to go about it.
Next, similarly, I thought that I could attempt a proof by contradiction, by assuming $U$ always preserves infinite limits, but this time working with the forgetful functor and aiming for a contradiction in $\mathbf{Sets}$.
To this end, I have some category $\mathbf{I}$ to function as the infinite diagram category that acts on both $\mathbf{B}G$ and $\mathbf{Sets}$. So consider
$$\begin{align} U_\mathbf{I}: (\mathbf{B}G)^\mathbf{I} & \longrightarrow\mathbf{Sets}^\mathbf{I}\\ ((X, \sigma)\stackrel{f}{\to}(Y, \tau))^\mathbf{I} & \stackrel{?}{\mapsto} (X\stackrel{f}{\to} Y)^\mathbf{I}. \end{align}$$
Does the map $U_\mathbf{I}$ have to satisfy some universal property or something with respect to $\mathbf{I}$? I'm quite lost here.
I'm not sure how to work with $((X, \sigma)\stackrel{f}{\to}(Y, \tau))^\mathbf{I}$ and $(X\stackrel{f}{\to} Y)^\mathbf{I}$ either, whatever they are.
I think what I have so far might even be nonsense.
Please help :)