I am trying to prove an extension to the result given as a Lemma below. I would appreciate feedback, and pointers to other ways of proving it.
Theorem. Let $V$ be a finite dimensional vector space. If the self-adjoint operators $A_1,\dots,A_r:V\to V$ commute, then there exists a simultaneous orthonormal eigenbasis for $V$ for $A_1,\dots,A_r$.
Lemma. Let $V$ be a finite dimensional vector space. If the self-adjoint operators $A_1,\dots,A_r:V\to V$ commute, then there exists a simultaneous eigenbasis for $V$, for $A_1,\dots,A_r$, though not necessarily orthonormal.
Proof of Theorem. We prove it by induction on $n$. Suppose first that $n=1$. Then there is a vector $e\in V$ with $||e|| = 1$, which is an ON-basis for $V$, and it is an eigenvector to all $A_1,\dots,A_r$. So the statement is true when $n=1$. Suppose that the statement is true also for $n=k$, and consider the case when $n=k+1$.
Let $e_1$ be a vector in the simultaneous eigenbasis that exists by the Lemma. Without loss of generality we may assume that $||e_1||=1$. Now let $U=(\text{span}\{e_1\})^{\bot}$. Denote by $\lambda_1^j$ the eigenvalue of $A_j$ to which the eigenvector $e_1$ corresponds. Since $A_j$ is self-adjoint, we have for an arbitrary vector $u\in U$ that $$ 0 = \langle \lambda_1^je_1,u\rangle = \langle A_je_1,u\rangle = \langle e_1,A_j u\rangle, $$ which shows that $U$ is invariant under the operators $A_1,\dots,A_r$. So the restrictions $A_1\rvert_U,\dots,A_r\rvert_U$ are self-adjoint (since $A_1,\dots,A_r$ are self-adjoint) linear operators on $U$.
By the induction hypothesis there exists a simultaneous ON-eigenbasis $\{e_2,\dots,e_{k+1}\}$ for $U$, since $\dim U = k$. But these vectors are also eigenvectors of $A_1,\dots,A_r$, and they are all orthogonal to $e_1$. Hence $e_1,\dots,e_{k+1}$ is a simultaneous ON eigenbasis for $V$. $\square$