Suppose $\mathbf A$ and $\mathbf B$ are both real, symmetric matrices. Suppose $\mathbf B$ has an eigenvalue $\lambda$ corresponding to eigenvector $\mathbf v$. Suppose that $\mathbf A$ and $\mathbf B$ commute. Therefore $$ \mathbf {BAv} = \mathbf {ABv} = \mathbf {A} \lambda \mathbf v = \lambda \mathbf {Av}$$ Therefore $\mathbf {Av}$ is an eigenvector of $\mathbf B$.
But this isn't good enough. For example, maybe (the transformation associated with) $\mathbf A$ maps $\mathbf v$ to some subspace orthogonal to $\mathbf v$. In that case $\mathbf {Av}$ is still an eigenvector of $\mathbf B$, but $\mathbf A$ does not simply scale $\mathbf v$ so $\mathbf v$ is therefore not an eigenvalue of $\mathbf A$.
What is missing from my proof to show that $\mathbf v$ is an eigenvector of $\mathbf A$ ?
You cannot show it because it is not true. Consider
$$ A = \operatorname{diag}(1,2,3), B = \operatorname{diag}(4,4,5). $$
Then $e_1 + e_2$ is an eigenvector of $B$ associated to the eigenvalue $4$ but $e_1 + e_2$ is not an eigenvector of $A$.
The point is that if $A,B$ are simulteneously diagonalizable (as happens in the above example), it means that one can find some basis $(v_1,\dots,v_n)$ of vectors that consists of eigenvectors both for $A$ and $B$. It does not mean that any basis of eigenvectors of $B$ is a basis of eigenvectors for $A$.