The proof is from Principles of Quantum Mechanics by Shankar. The theorem is:
If $\Omega$ and $\Lambda$ are two commuting Hermitian operators, there exists (at least) a basis of common eigenvectors that diagonalizes them both.
The proof is: Consider first the case where at least one of the operators is nondegenerate, i.e. to a given eigenvalue, there is just one eigenvector, up to a scale. Let us assume $\Omega$ is nondegenerate. Consider any one of its eigenvectors:
$$\Omega\left|\omega_i\right\rangle=\omega_i\left|\omega_i\right\rangle$$ $$\Lambda\Omega\left|\omega_i\right\rangle=\omega_i\Lambda\left|\omega_i\right\rangle$$ Since $[\Omega,\Lambda]=0$ $$\Omega\Lambda\left|\omega_i\right\rangle=\omega_i\Lambda\left|\omega_i\right\rangle$$
i.e., $\Lambda\left|\omega_i\right\rangle$ is an eigenvector with eigenvalue $\omega_i$. Since this vector is unique up to a scale,
Thus $\left|\omega_i\right\rangle$ is also an eigenvector of $\Lambda$ with eigenvalue $\lambda_i$...
What I do not understand is the statement/argument "Since this vector is unique up to a scale." I do not see how the argument allows to state the equation following it. What axiom or what other theorem is he using when he states "since this vector is unique up to a scale"?