 Solution

If the spectrum has one point, the preceding part gives $T=\lambda I$, and the linear span of any nonzero vector is a closed proper invariant subspace because $\dim H>1$.

Otherwise choose distinct points ${\lambda }_1,{\lambda }_2\in K$ and disjoint nonempty relatively open neighborhoods $U,V\subseteq K$. The open-set result gives $P(U)\ne 0$ and $P(V)\ne 0$, whileThus $P(U)\ne I$. Set $L=\mathrm{ran}P(U)$. As the range of an orthogonal projection, $L$ is closed, and the preceding observations give $\{0\}\ne L\ne H$. The Borel functional calculus for a normal operator gives $TP(U)=P(U)T$, soIn fact both $L$ and its orthogonal complement are invariant.