First, by Cauchy-Schwarz, we have for all $x \in H$ such that $\lVert x \rVert = 1$
\begin{equation*}
\lvert \langle Tx, x \rangle \rvert \leq \lVert Tx \rVert \underbrace{\lVert x \rVert}_{=1} \leq \lVert T \rVert
\end{equation*}
By the polarization identity, we have for all $(x, y) \in H^2$
$$
4\langle Tx, y \rangle
= \langle T(x + y), x + y \rangle - \langle T(x - y), x - y \rangle
+ i\langle T(x + iy), x + iy \rangle - \langle T(x - iy), x - iy\rangle.
$$
Applying triangle inequality we get
$$
4\lvert \langle Tx, y \rangle \rvert
\le w(T) \big[\lVert x + y\rVert^2 + \lVert x - y \rVert^2 + \lVert x + iy \rVert^2 + \lVert x - iy \rVert^2\big]
$$
Finally, by the identity $\lVert x + y\rVert^2 + \lVert x - y \rVert^2 + \lVert x + iy \rVert^2 + \lVert x - iy \rVert^2=4[\lVert x \rVert^2 + \lVert y \rVert^2]$:
\begin{equation*}
4\lvert \langle Tx, y \rangle \rvert \leq 4w(T)[\lVert x \rVert^2 + \lVert y \rVert^2]
\end{equation*}
Thus, if we take $\lVert x \rVert = \lVert y \rVert = 1$:
\begin{equation*}
\lvert \langle Tx, y \rangle \rvert \leq 2w(T)
\end{equation*}
As this is true for all $x, y \in H$ such that $\lVert x \rVert = \lVert y \rVert = 1$, you have: $\lVert T \rVert \leq 2w(T)$.
