If $F$ is an invariant subspace of $\Psi\colon G\to \UU(E)$ and $P\colon E\to F$ the orthogonal projection onto $F$. Then for all $g\in G$ the operators $P$ and $\Psi(g)$ are simultaneously diagonalizable.

We already know that $P$ commutes with $\Psi(g)$ for all $g\in G$. Hence, it suffices to show that $P$ and $\Psi(g)$ are diagonalizable because then by Proposition, $P$ and $\Psi(g)$ are simultaneously diagonalizable. We will show that $P$ and $\Psi(g)$ are normal operators: $P$ is self-adjoint (because it is an orthogonal projection), so it commutes with $P^*=P$ and is thus normal. By definition, $\Psi(g)$ is a unitary operator, so we have $\Psi(g)^*\Psi(g) = 1_E = \Psi(g)\Psi(g)^*$, and thus, $\Psi(g)$ is also normal. Hence, by Proposition, $P$ and $\Psi(g)$ are diagonalizable.