I have given the Schur's Lemma in following version:
Let $R:G \rightarrow \text{U}(\mathcal{H})$ be an irreducible representation of $G$ on $\mathcal{H}$. If $A \in \text{L}(\mathcal{H})$ satisfies
$$A R(g) = R(g) A \quad \forall g \in G$$
then $A = c I$ for some $c \in \mathbb{C}$.
Here $\mathcal{H}$ states a finite dimensional Hilbert space and $\text{U}(\mathcal{H})$ denotes the subspace of unitary operators. $R$ describes a homomorphism of a group $G$ on $\text{U}(\mathcal{H})$.
The proof I have given shows that it is sufficient to only prove this for hermitian $A$. So far so good. It continues with introducing a eigenvector $\left| \psi \right\rangle$ of $A$ so that the eigenspace of this operator is given by $\text{Eig}_\lambda(A) = \{\left| \psi \right\rangle: A\left| \psi \right\rangle = \lambda \left| \psi \right\rangle\}$. Then it states that $R(g) \left| \psi \right\rangle \in \text{Eig}_\lambda(A)$, because of $AR(g) \left| \psi \right\rangle = R(g) A \left| \psi \right\rangle = \lambda R(g) \left| \psi \right\rangle$. From that it is concluded that $\text{Eig}_\lambda(A)$ is an invariant subspace and because of $R$ being irreducible $\text{Eig}_\lambda(A)=\mathcal{H}$ follows.
My problem with this proof is now that I don't have a clue why you can conclude that $\text{Eig}_\lambda(A)$ is an invariant subspace of $R$ only because of the statement $A R(g)\left| \psi \right\rangle= \lambda R(g) \left| \psi \right\rangle$.