I found this exercise:
If $T$ is a compact operator and $S\in L(X)$ an invertible operator in $L(X)$ then there exists an $\alpha>1$ such that $S+\alpha T$ is invertible in $L(X)$.
Now the way I though about this is that we have $S+\alpha T=\alpha S(\frac{1}{\alpha}I+S^{-1}T)$, assume $\alpha\neq 0$, and we know that $S^{-1}T$ is a compact operator and since we are in a banach space we have that it's spectrum is $\{0\} \cup \sigma_p(S^{-1}T)$ where the latter is a countable set. And so we know that there exists $-\frac{1}{\alpha} <1$ such that $\frac{1}{\alpha}I+S^{-1}T$ is invertible. Now doesn't this say even more that there is an uncountable number of $\alpha's$ such that $\frac{1}{\alpha}I+ST^{-1}$ is invertible ? Or am I making some kind of mistake in the proof ? Thanks in advance.
What you have is fine. The proof breaks everything down to the fact that $\sigma(S^{-1}T)\cap(0,\infty)$ is countable (because $T$ is compact), while $(0,\infty)$ is uncountable, and you hit all those key points.