$\sigma_p(T)$ is at most countable for a normal operator

Question

This is the part (b) of Rudin's Functional Analysis Chapter 12, question 18.

Show that if $H$ is a separable Hilbert space then the point spectrum (denoted by $\sigma_p(T)$) of a normal operator $T$ is at most countable.

I have done part (a) and (c)
Part (a) is to prove that, the residual spectrum $\sigma_r(T)$ of a normal $T \in \mathcal{B}(H)$ is empty. (perhaps this result is useful)

Answer 1

If $\lambda_1,\lambda_2$ are distinct complex numbers in the point spectrum $\sigma_p(T)$, and if $e_1,e_2$ are unit vectors such that $Te_j=\lambda_j e_j$, then $\langle e_1,e_2\rangle = 0$. So, if you are working in a separable space, the point spectrum $\sigma_p(T)$ must be at most countable.