I am pretty sure that my problem has already discussed, but I didn't find.
So, the question is how to prove that two commuting operators have a common eigenvector.
The first note is following: Let $A$ be the operator such that $Av = \lambda v$ (we are considering algebraically closed case, so such $v$ surely exists). Then $ABv = BAv = \lambda Bv$, so $Bv$ is eigenvector of $A$ with eigenvalue $\lambda$ as well. But I am not sure, what to do next.
Best Answer
Let $\lambda$ an eigenvalue of $A$, and $E=\ker(A-\lambda \operatorname{Id})$.
Then $B(E) \subset E$ as you have shown.
So $B|_{E}$ has en eigenvector $x$ and an eigenvalue $\mu$ such that $B(x)=\mu x$.
$x \in E$ so $A(x)=\lambda x$ too.