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:
$$\sigma(N)\subseteq\overline{\mathcal{W}(N)}$$

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, $\lambda \in \rho(A)$.
