Proof of Schur lemma

The kernel and image of an intertwiner are invariant subspaces. Irreducibility therefore makes a nonzero intertwiner injective and surjective. For an endomorphism $T$ of a finite-dimensional irreducible complex representation, choose an eigenvalue $\lambda$. The noninvertible intertwiner $T-\lambda I$ must be zero, so $T=\lambda I$.