By your definition
$$
\|T\| = \sup\{ |\langle Tx,y\rangle| : \|x\|=\|y\|=1\},
$$
we can find sequences of $x_n,y_n\in H$ such that $\|x_n\|=\|y_n\|=1$ and $|\langle Tx_n,y_n\rangle| \to \|T\|$. More specifically, we can choose $x_n,y_n$ so that
$$
|\langle Tx_n,y_n\rangle| \ge \|T\| -\frac 1n.
$$
Now, consider the sequence $x'_n=\frac n{n+1}x_n, y'_n=\frac n{n+1}y_n$. It is obvious that $\|x'_n\|=\|y'_n\| <1$. Direct computation shows that
$$
|\langle Tx'_n,y'_n\rangle|=\frac {n^2}{(n+1)^2}|\langle Tx_n,y_n\rangle| \ge \frac {n^2}{(n+1)^2}\|T\| - \frac {n}{(n+1)^2}.
$$
By taking $n\to\infty$, we can see that
$$
\sup_{n\in\Bbb N} |\langle Tx'_n,y'_n\rangle| \ge \|T\|,
$$
and this is exactly what you need.
One can easily verify that $\|A\|=\sup\{|\langle Ax,y\rangle| : x,y \in \mathcal H,\
\|x\|=\|y\|=1\}$ in a complex Hilbert space $\mathcal H$.
Notice that $$\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle = 2\langle Ax,y\rangle+2\langle Ay,x\rangle.$$ Since $A$ is self adjoint, $\langle Ay,x\rangle=\langle y,Ax\rangle=\overline{\langle Ax,y\rangle}.$ So $$\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle=4\newcommand{\re}{\operatorname{Re}}\re\langle Ax,y\rangle.$$
Let $P:=\sup_{||x||\leq 1}|\langle x,Ax\rangle|.$ Then
$$\begin{align*}
|4\re\langle Ax,y\rangle|
&=|\langle A(x+y), x+y\rangle − \langle A(x- y), x- y\rangle |\\
&\leq P\|x+y\|^2+P\|x-y\|^2\\
&=2P(\|x\|^2+\|y\|^2).
\end{align*}$$
So, whenever $\|x\|=\|y\|=1$, we have $$|\re\langle Ax,y\rangle|\leq P\tag{$\color{red}{1}$}\label1.$$
Suppose $\langle Ax,y\rangle=re^{i\theta}$ with $\|x\|=\|y\|=1$. I will construct an element $z$ with $\|z\|=1$ such that $|\langle Ax,y\rangle|=|\re\langle Az,y\rangle|$. Then we can apply $(\ref 1)$ to $|\re\langle Az,y\rangle|$ and we will get $|\langle Ax,y\rangle|\leq P $.
Consider $z=e^{-i\theta} x$. Then $\|z\|=1$. Also, note that $$\langle Az,y\rangle=e^{-i\theta}\langle Ax,y\rangle=r=\re\langle Az,y\rangle,$$ and $|\langle Ax,y\rangle|=r$. So $$|\langle Ax,y\rangle|=r=|\re\langle Az,y\rangle|\leq P.$$ Hence $\|A\|\leq P$.
I learnt this proof from Functional Analysis by B.V. Limaye.
Best Answer
Yes, if the operator is self-adjoint. Here is a proof from Conway's book, A Course in Functional Analysis:
Let $M = \{\sup |\langle Ax, x\rangle| : \|x\| = 1\}$. If $\|x\| = 1$, then $|\langle Ax,x\rangle | \leq \|A\|$ (since this holds for all pairs of vectors, so certainly works if we use the same in each slot), so $M \leq \|A\|$.
Now let $h,g$ be unit vectors. Self-adjointness gives $$\langle A(h \pm g) , h \pm g\rangle = \langle Ah,h\rangle \pm 2\mbox{Re}\langle Ah,g\rangle + \langle Ag,g \rangle.$$ Subtracting these equations gives $$\langle A(h + g), h + g \rangle - \langle A(h-g), h-g \rangle = 4 \mbox{Re} \langle Ah,g \rangle.$$ Since $|\langle Af, f \rangle| \leq M\|f\|^2$ for any $f$, by Cauchy-Schwarz, we have using the parallelogram law that \begin{align*}&4\mbox{Re}\langle Ah,g\rangle \\\leq &M(\|h + g\|^2 + \|h-g\|^2) \\ =&2M(\|h\|^2 + \|g\|^2) \\ = &4M. \end{align*}
To finish up, choose $e^{i\theta}$ so that $\mbox{Re}(e^{i\theta}\langle Ah, g\rangle) = |\langle Ah,g \rangle|$. Then $$|\langle Ah, g \rangle| = \mbox{Re}(\langle Ae^{i\theta}h, g\rangle ) \leq M.$$ Take the supreumum over all $h,g$ to finish.