Bounded extension of an operator with limited numerical range

Question

In an excercise I'm asked to prove that a densely defined operator, whose numerical range:
$$\nu(T)=\{ (\psi,T\psi) \space | \space \psi \in D(T) \wedge ||\psi||=1 \}$$ is a limited subset of $\mathbb{C}$, admits a bounded extension, with the additional property that D(T)$\subset$D($T^{\ast}$). I thought that, taking the closure of the numerical range $\overline{\nu(T)}$, it is a closed and limited subset of $\mathbb{C}$ and so it is compact. Since D(T) is dense in the Hilbert space $\mathcal{H}$, every element $\phi\in\mathcal{H}\backslash D(T)$ can be seen as the limit of a sequence $\{\phi_n\}_{n\in\mathbb{N}}$ of elements in D(T). Based on this, every sequence in $\overline{\nu(T)}$ admits a converging subsequence. So $$ \exists\{\phi_{n_k}\}\subset D(T), K\in\overline{\nu(T)} \space | \space \lim_{n_k\rightarrow\infty}(\phi_{n_k},T\phi_{n_k})=(\phi,\lim_{n_k\rightarrow\infty}T\phi_{n_k})=K $$ (I took both $||\phi||=1$ and $||\phi_{n}||=1 \space \forall$ n beacuse otherwise it would only change the value of K, but it would still be a finite complex number)
This means that $\lim_{n_k\rightarrow\infty}T\phi_{n_k}$ exists and I can define T$\phi\doteq\lim_{n_k\rightarrow\infty}T\phi_{n_k}$.
Is this enough to conclude that in this way i defined a bounded extension of my operator?