Suppose that $T$ is a bounded operator with finite spectrum. What happens with the spectrum of $T+F$, where $F$ has finite rank? Is it possible that $\sigma(T+F)$ has non-empty interior? Is it always at most countable?
Update: If $\sigma(T)=\{0\}$ then $0$ is in the essential spectrum of $T$ ($T$ is not invertible in the Calkin algebra), hence for any compact $K$, $\sigma_{ess}(T+K)=\sigma_{ess}(T)=\{0\}$. For operators such that the essential spectrum is $\{0\}$, it is known that their spectrum is either finite or consists of a sequence converging to $0$. I think it should be the same for operators with finite spectrum, but I cannot find a proof or reference.
It is always true if $T$ is self-adjoint. Here is a theorem that you might be interested:
If $T$ is self-adjoint, a complex number is in the spectrum of $T$ but not in its essential spectrum iff it is an isolated eigenvalue of $T$ of finite multipliticity.
The result can be found on page 32 of Analytic K-homology by Nigel Higson and John Roe.