Normal Operators: Numerical Range Disclaimer:
As I realized in the comments that this works for normal operators I decided to modify this question. Besides, I got the proof now - thanks to T.A.E.!
Prove that for normal operators the spectrum is contained in the closure of the numerical range:
$$\sigma(N)\subseteq\overline{\mathcal{W}(N)}$$
 A: If the distance from $\lambda$ to $\mathcal{W}(A)$ is $d > 0$, then
$$
              |((A-\lambda I)\phi,\phi)| \ge d\|\phi\|^{2},\;\;\; \phi \in \mathcal{D}(A)\\
               \implies d\|\phi\| \le \|(A-\lambda I)\phi\|,\;\;\; \phi \in\mathcal{D}(A).
$$
If $(A\phi,\phi)=(\phi,A^{\star}\phi)$ for $\phi$ in a core domain of $A^{\star}$, then you also get
$$
              d\|\phi\| \le \|(A^{\star}-\overline{\lambda}I)\phi\|,\;\;\;\phi \in \mathcal{D}(A^{\star}).
$$
So, in this case $\mathcal{N}(A^{\star}-\overline{\lambda}I)=\{0\}$. Using the adjoint relation, $\mathcal{R}(A-\lambda I)^{\perp}=\mathcal{N}(A^{\star}-\overline{\lambda}I)$, it follows that $(A-\lambda I)$ has dense range and bounded inverse. So, under these assumptions, $\lambda \in \rho(A)$.
A: Disclaimer: This prove is mostly due to T.A.E. - greatest regards to him!

Preparation:
(This is where real work has to be done.)
The distance of $\lambda$ to the numerical range gives us the estimate:
$$d(\lambda,\mathcal{W})\leq|\langle \hat{x},(N-\lambda)\hat{x}\rangle|\leq\|(N-\lambda)\hat{x}\|\implies \|(N-\lambda)x\|\geq d(\lambda,\mathcal{W})\|x\|$$
Moreover since $N$ is normal and densely defined we have $N^*N=NN^*$ as a really nontrivial result (see proposition 4.17 b in [1]). Also since $N$ is normal it is closed and since it is densely defined $\mathcal{D}(NN^*)$ is a core for $N^*$ (see proposition 4.11 a in [1]). We thus have:
$$d(\lambda,\mathcal{W})\leq|\langle \hat{c},(N-\lambda)\hat{c}\rangle|=|\langle (N^*-\overline{\lambda})\hat{c},\hat{c}\rangle|\leq\|(N^*-\overline{\lambda})\hat{c}\|\quad \hat{c}\in\mathcal{D}(N^*N)=\mathcal{D}(NN^*)\implies\|(N^*-\overline{\lambda})y\|=\lim_n\|(N^*-\overline{\lambda})c_n\|\geq\lim_n d(\lambda,\mathcal{W})\|c_n\|=d(\lambda,\mathcal{W})\|y\|$$
Main Work:
(The prove will be now an easy consequence.)
Assume $\lambda\notin\overline{\mathcal{W}}$.
Since $\lambda$ has positive distance to the numerical range $(N-\lambda)$ will be bounded below by the first estimate given in the preparation and due to this especially injective. Moreover since $N$ is normal it is closed so as well $(N-\lambda)$ and $(N-\lambda)^{-1}$. Thus by the closed graph theorem applied to $(N-\lambda)^{-1}$ the range of $(N-\lambda)$ is closed.
On the other hand we have that the $(N^*-\overline{\lambda})$ is bounded below as well due to the second estimate given in the preparation. But therefore the kernel is zero only and thus we have:
$$\overline{\mathcal{R}(N-\lambda)}=\mathcal{N}(N^*-\overline{\lambda})^\perp=\{0\}^\perp=Y$$
That is the range of $(N-\lambda)$ is dense.
Collecting all together $(N-\lambda)$ is injective surjective and bounded below.
So $\lambda$ was in the resolvent set as was to be shown.

Reference: [1]: German version of Weidmann's 'Lineare Operatoren in Hilberträumen'
