A slightly pedantic comment: the group $G$ in Mark's posting does not come equipped with an action on $\mathbb{R}^5$. The manifold $M'$ with the proper $G$-action is constructed from $G\times M$, i.e., the direct product of $G$ and $M$, by identifying pieces of the various copies of $\partial M$. Provided that the 2-complex $C$ that you started with was aspherical, the universal cover of $M'$ will be contractible. But even in this case, it need not be homeomorphic to $\mathbb{R}^5$.

Translations and glides in all directions might be a union of conjugacy classes, so in that sense it's normal, but it isn't a subgroup! It generates the full group of isometries of the plane.

The group $U=K*_GH$ does have the required property, and the Mayer-Vietoris sequence proves this. What did you think was the problem with proving this using Mayer-Vietoris?

In question 2, do you want each $G_n$ to be connected? If not, then surely any profinite group is a closed subgroup of a Tychonoff product of finite groups?

Of course, the mapping telescope of the system of maps that Dan describes is a model for $K(\mathbb{Q},1)$. One way to look at this example is as a case where hocolim is not homotopy equivalent to colim.

An action on a cube complex that is not finite-dimensional can never be cocompact. The cube complex that Farley uses is not finite-dimensional.