Eigenvalues of a compact operator with modulus greater than $\epsilon$ is finite

Question

Consider the following fragments from Murphy's "$C^*$-algebras and operator theory":

In the above proof of theorem 2.4.5., how does theorem 1.4.11 imply that the set $S$ is finite? I guess it has something to do with the non-zero points of $\sigma(u)$ being isolated?

Answer 1

Let $x_n$ be a sequence in $S$. Since $S$ is bounded in $\Bbb C$ it is a bounded sequence and as such admits a convergent subsequence, so assume it converges. Further $|x_n|≥\epsilon$ for all $n$ so $x_n$ cannot converge to $0$. Since $S=\sigma(u)\cap \{ x\in\Bbb C\mid |x|≥\epsilon \}$ is closed the limit must also lie in $S$. But every point of $S$ is isolated and as such cannot be the limit of other points and $x_n$ must be eventually constant (ie $S$ is discrete in $\Bbb C$).

This implies $S$ is finite, for otherwise there must be a sequence without a constant subsequence.