I'm working on understanding a proof that if an operator $A$ on a Hilbert space $\mathcal{H}$ is compact, then show that $\sigma(A) - \{0\} \subseteq \sigma_p(A)$. If you're not familiar with this notation, $\sigma(A)$ is the spectrum and $\sigma_p(A)$ is the point spectrum. The first part of the proof succinctly demonstrates how the approximate point spectrum $\sigma_{ap}(A) \subseteq \sigma_p(A)$ and now the proof goes on to claim that $\sigma_p(A)$ is countable. This is the part of the argument I don't understand and would like some clarification. Here is the argument paraphrased:
Let $\lambda_1, \lambda_2, \lambda_3, \ldots$ be distinct eigenvalues of $A$ with associated eigenvectors $f_1, f_2, f_3, \ldots$ where $f_n \neq 0$. Let $g_1, g_2, \ldots$ define an orthonormal set of vectors such that for all $n \in \mathbb{N}, \; g_n \in Span\{f_1, \ldots, f_n\}$. Observe that $Ag_n - \lambda_ng_n \in Span\{f_1, \ldots, f_{n-1}\}$ thus we can express $\lambda_n = \langle Ag_n, g_n\rangle$. Since $A$ is compact we additionally have that $\lambda_n \to 0$ as $n \to \infty$.
How on earth is this argument sufficient for proving $\sigma_p(A)$ is countable? Each step of the argument is easy enough to understand but how does the main result follow?