Consider the following fragment:
Questions:
(1) Does every non-zero eigenvalue occur in the $\{\lambda_i: i =1, 2, \dots\}$?
(2) Why if $\lambda \ne 0$ is the eigenspace of $\lambda$ finite dimensional? I can see this must be the case if $(1)$ is true.
(1) this is the statement of the spectral theorem. Assume one eigenvalue $\tilde \lambda$ is missing in the enumeration in the theorem. Then the associated eigenvector $\tilde v$ is orthogonal to all the other eigenvectors that are used in the spectral theorem. This implies $\tilde v \in \ker T^{\perp\perp}=\ker T$. Hence $\tilde \lambda=0$, contradiction.