I suspect that if $G$ is a compact Lie group, that $H^2_\textrm{group}(G;U(1))$ is finite. (For example: $H^2_\textrm{group}(SO(3);U(1)) = \mathbb Z_2$)
I couldn't easily find this statement though, and I can't quite prove it. I think the following should be correct: $$H^2_\textrm{group}(G;U(1)) \cong H^2(BG;U(1)) \cong H^2(G;\mathbb Z)$$
Hence, what I want to show is equivalent to showing that the singular cohomolgoy $H^2(G;\mathbb Z)$ is purely torsion. Or again, equivalently, that $H^2_\textrm{dR}(G) \cong H^2(G;\mathbb R) = 0$.
Can anyone confirm this, or give a counter-example?
Just to post the comment as an answer. We will provide a counterexample. Consider the torus $S^1\times S^1$. It is a compact Lie group since it is the product of two compact Lie groups. However, by Poincaré duality (or computing it directly), we have $H^2(S^1\times S^1,\mathbb{Z})\cong \mathbb{Z}$.