 Solution

Let $Av=\lambda v$ with $v\ne 0$. A vanishing commutator gives $A(Bv)=B(Av)=\lambda Bv$, so $Bv$ belongs to the eigenspace $E_{\lambda }$. This is eigenspace invariance for commuting linear maps. A nondegenerate eigenvalue has a one-dimensional eigenspace, so $Bv=\beta v$ for a scalar $\beta$; $\beta =0$ is allowed. Therefore $v$ is also an eigenvector of $B$. Without the simple-eigenvalue condition, $B$ can mix vectors inside a larger eigenspace, and an arbitrarily chosen eigenvector of $A$ need not be an eigenvector of $B$.