Inferences made from $(T-2I)(T-3I)(T-4I)v = 0$ Is the following Proof of the given theorem Correct?

Given that $T\in\mathcal{L}(V)$ and $(T-2I)(T-3I)(T-4I) = 0$ and
  $\lambda$ is an eigenvalue of $T$ then $\lambda = 2$ or $\lambda = 3$
  or $\lambda = 4$.

Proof. We prove the equivalent claim that if neither $2$ nor $3$ are eigenvalues of $T$ then $4$ must be. Assume that $\lambda\neq 2$ and $\lambda\neq 3$ and let $v$ be the eigenvector corresponding to $\lambda$. Since $(T-2I)(T-3I)(T-4I) = 0$ then in particular for $v$ we have 
$(T-2I)(T-3I)(T-4I)v = 0$ which implies that at least one of the operators $(T-2I)$ or $(T-3I)$ or $(T-4I)$ is not injective, from hypothesis $2$ and $3$ are not eigenvalues of $T$ consequently $(T-2I)$ and $(T-3I)$ are injective therefore for some non-zero $w\in V$ $(T-4I)w = 0$ implying $\lambda = 4$. 
$\blacksquare$
 A: It's very simple to do this directly: Suppose $\lambda$ is an eigenvalue of $T$. Choose $v\ne0$ such that $Tv=\lambda v$. Then $$0=(T-2I)(T-3I)(T-4I)v=(\lambda-2)(\lambda-3)(\lambda-4)v,$$hence $(\lambda-2)(\lambda-3)(\lambda-4)=0$.
In general, if $p$ is a polynomial, $p(T)=0$, and $\lambda$ is an eigenvalue of $T$ then $p(\lambda)=0$: If $v\ne0$ and $Tv=\lambda v$ then $$0=p(T)v=p(\lambda)v.$$
A: No, this is not correct.  Your "equivalent claim" is not actually equivalent.  For instance, what if the set of eigenvalues of $T$ is $\{2,5\}$?  Then the statement you're trying to prove would be false (since $5$ is not supposed to be an eigenvalue), but your "equivalent claim" is true (since $2$ is an eigenvalue so the antecedent of the implication is false).
(What would be an equivalent claim is that "for any eigenvalue $\lambda$, if $\lambda\neq 2$ and $\lambda\neq 3$, then $\lambda=4$".  Your version moves the quantifier on $\lambda$ inside to the individual statements instead of having a single quantifier outside the whole statement.)
