Prove that if $\{b_n\}$ are eigenvectors with distinct eigenvalues $\{\lambda_n\}$, then $\{b_n\}$ are linearly independent.
Note: Proofs for this are available on this site and elsewhere. My question is to verify my proof below, which, at least to me, is simpler.
Lemma: If $B$ is a finite set of linear independent vectors, there is at most one linear combination of $B$ equal to a given vector $v$.
Proof: Consider a finite set $\{b_n\}$ of linearly independent vectors. Assume $\sum_n c_nb_n = v$ and $\sum_n c'_nb_n = v$, with at least one $c_n \neq c'_n$. Then $\sum_n (c_n - c'_n)b_n = v - v = 0$, which is impossible since $\{b_n\}$ is linearly independent.
Main Proof: Let $\{b_n\}$ be linearly independent eigenvectors of matrix $A$ with distinct eigenvalues $\{\lambda_n\}$. Let $d$ be a distinct eigenvector of $A$ with eigenvalue $\delta \notin \{\lambda_n\}$. We claim that $d$ is linearly independent of $\{b_n\}$.
Assume $d$ is not linearly independent of $\{b_n\}$. Then there exists $\{c_n\}$ such that $\sum_n c_nb_n = d$. Consequently, $Ad = \sum_n \lambda_n c_n b_n$. But $Ad = \delta d = \sum_n \delta c_n b_n$. This would give two different linear combinations of $\{b_n\}$ that yield $Ad$, violating the lemma. Hence, $d$ must be linearly independent of $\{b_n\}$.
Questions: Is this proof correct, rigorous, and well written? How can it be improved?
As far as I can tell, this proof does not use induction. Is that correct?
Update
Thank you the responders. I believe I need to add the following step, which indeed uses induction. Can you please verify that with this step, the proof is correct?
Let $B_0$ be the empty set, and for all $n \in \mathbb N$, let $B_n = B_{n-1} \cup \{b_n\}$. If $B_n$ is a set (possibly empty) of linearly independent eigenvectors with distinct eigenvalues, $B_{n+1}$ must also be, as proven above. $B_0$ is such a set, and so therefore $B_n$ is such for all $n$. This completes the proof.