Consider the following proof in Murphy's book: "$C^*$-algebras and operator theory":
I try to understand why $u(K^\perp) \subseteq K^\perp$. Equivalently, one can prove $u^*(K) \subseteq K$.
I tried to calculate $\langle u(x), y \rangle$ with $x \in K^\perp, y \in K$ and I hoped to calculate that this is equal to $0$ but I must be missing something.
At first observe that if $u$ is normal and $u(x) = \lambda x$, then $u^*(x) = \overline{\lambda} x$ (it is a simple exercise).
Now we can prove the following proposition: if $u$ is normal, $E$ is arbitrary set of eigenvectors of $u$, and $K = \overline{span(E)}$, then $K^\perp$ is $u$-invariant. Indeed, $x \in K^\perp$ iff $\langle x, a \rangle = 0$ for all $a \in E$. Thus, for all $a \in E$, $\langle u(x), a \rangle = \langle x, u^*(a) \rangle = 0$ (by the foregoing observation). Therefore, $u(x) \in K^\perp$.
$\mathbf{Edit}$: proof of the observation.
It is obvious that operator $v = u - \lambda I$ is normal, since $v^* = u^* - \overline{\lambda}I$. Let $v(x) = 0$. Then $$ 0 = \langle v(x), v(x) \rangle = \langle v^*(v(x)), x \rangle = \langle v(v^*(x)), x \rangle = \langle v^*(x), v^*(x) \rangle = 0$$ Thus, $v^*(x) = 0$, i.e. $u^*(x) = \overline{\lambda}(x)$.