In Normal $T\in B(H)$ has a nontrivial invariant subspace, Haskell Curry said (for case (ii)) to pick two open, disjoint subsets of $\sigma(T)$ where $|\sigma(T)|\geq 2$. I don't think that this is always possible and some smaller fixes also didn't finish the job. The problem:
If the spectrum only contains countably many elements, no empty subset is open (given our usual topology).
Furthermore, if we had any non-empty open subset, we should be finished, as we can find disjoint, open balls inside and use these. So my question really reads:
Can we find two disjoint sets $A,B\subset \sigma(T)$ with $E(A)\neq0\neq E(B)$ if the spectrum only contains its boundary $\partial\sigma(T)$ and no interior points?
First off, we can ignore isolated points $\lambda$ since they are eigenvalues and the argument still works as $E(\{\lambda\})\neq 0$. Similarly, if $T$ has at least $2$ eigenvalues, we are finished. I had hoped that $(\sigma(T))^o=\emptyset$ would imply $\sigma_p(T)=\partial\sigma(T)=\sigma(T)$ (similar to compact operators) but I couldn't find any evidence and in general, we definitely can't say $\partial\sigma(T)\subseteq\sigma_p(T)$ (as some operators don't even have any eigenvalues).
Any help would be appreciated.
EDIT: I just had another idea. For any (Borel-)subset $A\subseteq\sigma(T)$, we know $$1=E(\sigma(T))=E(A)+E(\sigma(T)\setminus A)$$ Now let's set $B:=\sigma(T)\setminus A$. If $E(A)\notin\{0,1\}$, then $E(B)\notin\{0,1\}$ and we have found our two subsets. Otherwise, as $E(A)=0$ implies $E(B)=1$, we can restrict ourselves to the third and last case of $E(A)=1$. Now we can repeat this game and should find some subset $A_0$ with $E(A_0)\notin\{0,1\}$. If not, then we have $E(A)\in\{0,1\}$ for all (Borel-)subsets of $\sigma(T)$. Intuitively, it's obvious that the spectrum can only contain a single value $\lambda$ but I'm not quite sure how to show that.
Perhaps it is more transparent to apply the spectral theory of self adjoint operators. Assume $T$ is a normal operator and is is not a multiple of the identity operator. We have $T=A+iB$ where $A={1\over 2}(T+T^*)$ and $B={1\over 2i}(T-T^*).$ Then $A$ and $B$ are self-adjoint and $AB=BA.$ Moreover $A$ or $B$ is not a multiple of the identity operator. WLOG we may assume that $A$ is not a multiple of $I.$ Then $\sigma(A)$ contains at least two points $\lambda_0<\lambda_1.$ Let $E(t)$ denote the resolution of the identity associated with $A.$ Then $A$ and $B$ commute with $E(t).$ For $\lambda={1\over 2}(\lambda_0+\lambda_1)$ the operator $E(\lambda)$ is a nontrivial orthogonal projection such that $AE(\lambda)=E(\lambda)A$ and $BE(\lambda)=E(\lambda)B.$ Therefore the range of $E(\lambda)$ is a nontrivial closed invariant subspace for both $A$ and $B,$ hence also for $T.$