If $S\circ T=T\circ S$, then $S$ and $T$ have a common eigenvector.

If $S\circ T=T\circ S$, then $S$ and $T$ have a common eigenvector.

Assume $S$ and $T$ are diagonalizable maps on $\mathbb{R}^n$ such that
$S\circ T$=$T \circ S$. Then $S$ and $T$ have a common eigenvector.
I already have proof, but I just need validation in one part. My proof:
Let $F$ be an eigenvector of $T$. This means $\exists \; \lambda \in R$
such that $T(v)=\lambda v$. Then, using the fact that $S\circ T$=$T \circ
S$, we have
$$ S(T(v)) = (S\circ T)(v)=(T \circ S)(v)=T(S(v)) \Longrightarrow
T(S(v))=\lambda S(v)$$
Thus, $S(v)$ is also an eigenvector of $T$. So, $S$ maps eigenvectors of
$T$ to eigenvevtors of $T$. Thus, $S$ must have an eigenvector of $T$.
How would one rigorously prove that if $S$ maps eigenvectors of $T$ to
eigenvectors of $T$, then $S$ also has an eigenvector of $T$?
Thanks.