 Solution

The sequence of norms is bounded. Choose $M>{\sup }_n\Vert x_n\Vert$ and $M>\Vert x\Vert$. By the spectral radius formula, every relevant spectrum lies in $|\lambda |<M$. The compact set $K=\{\lambda :|\lambda |\le M\}\setminus U$ is disjoint from $\sigma (x)$. If nonempty, continuity of the resolvent of an element gives $C={\sup }_K\Vert (\lambda 1-x{)}^{-1}\Vert <\infty$. For large $n$, $C\Vert x_n-x\Vert <1$, andis invertible for every $\lambda \in K$ by the Neumann series. If $K$ is empty there is nothing to exclude. No spectral points can lie outside the radius-$M$ disc either. Consequently $\sigma (x_n)\subset U$ for all sufficiently large $n$. Taking $U$ to be the disc of radius $r(x)+\epsilon$ also proves upper semicontinuity of the spectral radius.