 Solution

Hermitian means $T=T^{\ast }$, unitary $T^{\ast }T=T{T}^{\ast }=I$, and normal $T^{\ast }T=T{T}^{\ast }$. If $\Vert (T-\lambda )x_n\Vert \to 0$ for unit $x_n$, then $⟨T{x}_n,x_n⟩\to \lambda$ and each term is real, so $\lambda \in \mathbb{R}$. For self-adjoint $T$, a spectral value that is not approximate would make $T-\lambda$ bounded below; its closed range has orthogonal complement $\ker (T-\bar{\lambda })=0$, hence is onto, contradicting spectrality.