For every subgroup $H\subset G$, $BH$ occurs as a covering space of $BG$. If $BG$ were finite-dimensional then every covering space would be finite-dimensional. But for $C_p$ cyclic of prime order $p$ the space $BC_p$ has nontrivial mod $p$ cohomology in infinitely many (in fact all) dimensions. This can be made pretty geometric: there is a nice cell structure with one cell in every dimension and manifolds (lens spaces) as the odd-dimensional skeleta ...

EDIT By the way, this also yields the more general statement that $BG$ cannot be finite-dimensional unless $G$ is torsion-free.

EDIT In response to a comment here are some details: Make $C_p$ act on $S^{2n-1}$, the unit sphere in $\mathbb C^n$, freely by $p$th roots of $1$. This sphere is $(2n-2)$-connected and the union as $n\to\infty$ is contractible, so the orbit space is a model for $BC_p$. One can describe a cell structure in $S^{2n-1}$ with $p$ cells in every dimension up to $2n-1$ yielding a cell structure on the orbit space with one cell in every dimension up to $2n-1$, so that $BC_p$ gets one cell in every dimension. You can work out the boundary maps and see that the mod $p$ cohomology is nontrivial in all dimensions. Or you can save some trouble by using Poincare duality, since these odd-numbered skeleta are manifolds.