If a representation $\Psi\colon G\to \UU(E)$ has an invariant subspace $F$ then for all $g\in G$ the operator $\Psi(g)$ must have an eigen-vector in $F$.

Let $g\in G$. Since $F$ is an invariant subspace of $\Psi$, we have $\Psi(g)(F) = F$. In particular, the image of the restriction $\Psi(g)|F$ is a subset of $F$, i.e. we have $\Psi(g)|F\colon F\to F$. Now since $F$ is a finite-dimensional, complex vector space, the operator $\Psi(g)|F$ has an eigen-vector in $F$, which is also an eigen-vector of $\Psi(g)$ in $F$.