If $\lambda$ is isolated in $\sigma(u)$, then $E(\left\{\lambda\right\})(H)=\ker(u-\lambda)$. This is a Question 2.11 from Murphy's book: C$^*$-algebras and Operator Theory: Let $H$ be a Hilbert space.

Let $u\in B(H)$ be a normal operator with spectral resolution of the identity $E$.
  (a) Show that $u$ admits an invariant closed vector subspace other than $0$ and $H$ if $\dim(H)>1$.
  (b) If $\lambda$ is an isolated point of $\sigma(u)$, show that $E(\lambda)=\ker(u-\lambda)$ and that $\lambda$ is an eigenvalue of $u$.

In item (b), I understand that we have to prove that $E(\left\{\lambda\right\})(H)=\ker(u-\lambda)$, so $E(\left\{\lambda\right\})$ is the projection over it.
I managed to solve item (a) (see here, for example), and some parts of item (b):
Let $\lambda$ be an isolated point of $\sigma(u)$. Then the characteristic function $\chi_\lambda:\mu\in\sigma(u)\mapsto\delta_{\mu,\lambda}\in\mathbb{C}$ is continuous and $(z-\lambda)\chi_\lambda=0$, where $z:\sigma(u)\hookrightarrow\mathbb{C}$ is the inclusion. Using the (continuous) functional calculus at $u$, we obtain $$(u-\lambda)E(\left\{\lambda\right\})=(u-\lambda)\chi_\lambda(u)=0,\qquad(*)$$since $\chi_\lambda(u)=E(\left\{\lambda\right\}$). This shows that  $E(\left\{\lambda\right\})(H)\subseteq\ker(u-\lambda)$. Also, $\left\{\lambda\right\}$ is a (non-empty) open subset of $\sigma(u)$, so $E(\left\{\lambda\right\})\neq 0$ (this is obtained using the Borel functional calculus), hence $\lambda$ is an eigenvalue of $u$.
Finally, what I don't know is how to show the inclusion $\ker(u-\lambda)\subseteq E(\left\{\lambda\right\})(H)$ or, equivalently, $E(\left\{\lambda\right\})(H)\subseteq\overline{(u^*-\overline{\lambda})(H)}$. If I knew that the projection of $H$ over $\ker(u-\lambda)$ belongs to $C^*(u)$, then an easy application of the continuous functional calculus at $u$ would give the result.
Finally, if we change, without loss of generality, $u$ by $u-\lambda$, we have to do the following:

Show that if $u$ is normal and $0$ is an isolated point of $\sigma(u)$, then $E(\left\{0\right\})(H)=\chi_0(u)(H)\subseteq\overline{u^*(H)}$.

(PS: This result is given here in the case that $u$ is hermitian, but the solution seem to be overcomplicated...)
 A: There is a more general result that can be found in Rudin's Functional Analysis in the chapter on bounded operators on a Hilbert space.
Suppose $T\in\mathcal{B}(H)$ is normal and $E$ is its spectral decomposition. If $f\in C(\sigma(T))$ and if $\omega_{0}=f^{-1}(0)$, then $\mathcal{N}(f(T))=\mathcal{R}(E(\omega_{0}))$ where $\mathcal{N}(f(T))$ denotes the null space of the operator $f(T)$ and $\mathcal{R}(E(\omega_{0}))$ denotes the range of the operator $E(\omega_{0})$.
The proof is as follows:
Let \begin{equation}
g(\lambda)=\begin{cases}
1, & \text{if $\lambda\in\omega_{0}$}.\\
 0, & \text{on other points of $\sigma(T)$}.
\end{cases}
\end{equation}
 Then $fg=0$ hence by the functional calculus, $f(T)g(T)=0$. But $g(T)=E(\omega_{0})$, hence $\mathcal{R}(E(\omega_{0}))\subset\mathcal{N}(f(T))$.
For the other inclusion:
For each $n\in \mathbb{N}$ let $\omega_{n}=\{\lambda\in \sigma(T):\frac{1}{n}\leq|f(\lambda)|<\frac{1}{n-1}\}$. The complement $\tilde{\omega}$ of $\omega_{0}$ relative to $\sigma(T)$ is then the union of the disjoint Borel sets $\omega_{n}$. Define 
\begin{equation}
f_{n}(\lambda)=\begin{cases}
\frac{1}{f(\lambda)}, & \text{on } \omega_{n}\\
0, & \text{on other points of $\sigma(T)$}
\end{cases}
\end{equation}
Then each $f_{n}$ is a bounded Borel function on $\sigma(T)$ and $f_{n}(T)f(T)=E(\omega_{n}) \forall n \in \mathbb{N}$.
So $f(T)x=0 \Rightarrow E(\omega_{n})x=0$. By the countable additivity of the $H$ valued map $\omega\rightarrow E(\omega)x$, we have $E(\tilde{\omega})x=0$. Hence $E(\omega_{0})x=x$.
Thus $\mathcal{N}(f(T))\subset \mathcal{R}(E(\omega_{0}))$.
For your question, consider the continuous function $f(z)=z-\lambda$.
