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

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 $S=sigma(u)cap { xinBbb Cmid |x|≥epsilon }$ 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.