If $V = \text{null}(\textsf{T}-\lambda\textsf{I}) \oplus \text{range}(\textsf{T}-\lambda\textsf{I})$, then prove that $\textsf{T}$ is diagonalizable

Let us enumerate the eigenvalues $\lambda_1,\cdots \lambda_k$. As eigenspaces corresponding to different eigenvalues have trivial intersection, we have that $\text{null}(T-\lambda_{i+1} I) \subset \text{range} (T-\lambda_i I) $ for each $i<k$. By induction and the condition given by the problem, this gives us $V=\bigoplus_{i=1}^k \text{null} (T-\lambda_i I) \,\oplus \,\bigcap_{i=1}^k \text{range}(T-\lambda_i I)$. Thus, it remains to show that the intersection of ranges is trivial. Observe that $W:=\,\bigcap_{i=1}^k \text{range}(T-\lambda_i I)$ is an invariant subspace of T. If $\dim W>0$, then $T|_W$ has an eigenvector, which is not possible since all the eigenspaces of $T$ are accounted for in $\bigoplus_{i=1}^k \text{null} (T-\lambda_i I)$, so we must have $W=\{0\}$ as required. It follows that $V$ decomposes into a direct sum of eigenspaces of $T$, so $T$ is diagonalizable.