$\|S\|=\sup\{|\langle Sx\;,\;y\rangle |;\;\|x\| \leq 1,\, \|y\| \leq 1\}\,?$ Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$. Let $S\in \mathcal{B}(F)$. The norm of $S$ is defined us
$$\|S\|:=\sup_{\substack{x\in F\\ x\not=0}}\frac{\|Sx\|}{\|x\|}$$
Why
$$\|S\|=\sup_{\substack{\|x\| \leq 1,\\ \|y\| \leq  1}}|\langle Sx\;,\;y\rangle |?$$
Thanks.
 A: We have first $\|S\|=\sup_{\|x\|\leq 1}\|Sx\|$. Now $\|Sx\|=\sup\{|\left<Sx,y\right>|: \|y\|\leq 1\}$.
Indeed, for $x\ne 0$, then $y:=\dfrac{x}{\|x\|}$ is such that $\|y\|=1$, then $\dfrac{\|Sx\|}{\|x\|}=\left\|S\left(\dfrac{x}{\|x\|}\right)\right\|=\|Sy\|\leq\sup_{\|x\|\leq 1}\|Sx\|$. So $\|S\|\leq\sup_{\|x\|\leq 1}\|Sx\|$.
Now, for $\|y\|\leq 1$, then $|\left<Sx,y\right>|\leq\|Sx\|\|y\|\leq\|Sx\|$, so $M:=\sup\{|\left<Sx,y\right>|: \|y\|\leq 1\}\leq\|Sx\|$. 
Now, $\left|\left<Sx,\dfrac{Sx}{\|Sx\|}\right>\right|=\|Sx\|$, and $\left\|\dfrac{Sx}{\|Sx\|}\right\|=1$, of course, if $Sx=0$, this leads to the trivial case.
Additional notes. If $M$ is a subspace of $F$ such that the image, $\text{Im}(S)$ is such that $\text{Im}(S)\subseteq\overline{M}$, in this case, we have 
\begin{align*}
\|S\|=\sup\{|\left<Sx,y\right>|: \|x\|\leq 1, \|y\|\leq 1, y\in M\}.
\end{align*}
Indeed, as $\dfrac{Sx}{\|Sx\|}\in\text{Im}(S)$, find a sequence $(y_{n})$, $y_{n}\in M$ such that $y_{n}\rightarrow\dfrac{Sx}{\|Sx\|}$. Now, set $z_{n}=\dfrac{y_{n}}{\|y_{n}\|}$. We have $\|y_{n}\|\rightarrow 1$ and hence $z_{n}\rightarrow\dfrac{Sx}{\|Sx\|}$ and that $\left<Sx,z_{n}\right>\rightarrow\left<Sx,\dfrac{Sx}{\|Sx\|}\right>=\|Sx\|$, but $\|z_{n}\|=1$, by the inequality that
\begin{align*}
|\left<Sx,z_{n}\right>|\leq\sup\{|\left<Sx,y\right>|: \|x\|\leq 1, \|y\|\leq 1, y\in M\},
\end{align*}
we deduce that
\begin{align*}
\|Sx\|\leq\sup\{|\left<Sx,y\right>|: \|x\|\leq 1, \|y\|\leq 1, y\in M\}.
\end{align*}
A: For a fixed $x$, we have $$\sup_{\|y\| \le 1} |\langle Sx, y \rangle| = \|Sx\|$$
by Cauchy-Schwarz (with supremum attained by $y = Sx/\|Sx\|$.
Thus
$$\sup_{\substack{\|x\| \le 1 \\ \|y\|\le 1}} |\langle Sx, y \rangle|
= \sup_{\|x\| \le 1} \|Sx\| = \|S\|.$$

If the last equality is not obvious to you, note
$$\sup_{\|x\| \le 1} \|Sx\| = \sup_{\|x\| = 1} \|S x\|$$
because if $\|x\| < 1$, then $\|Sx\| < \|S (x / \|x\|)\|$ so the supremum on the left-hand side must be attained by some $x$ with norm $1$.
Then, it is easy to show that the right-hand side is equivalent to your definition $\sup_{x \ne 0} \|Sx\| / \|x\|$.
A: For any $x, y \in X$ with $\|x\|, \|y\| \le 1$ we have
$$|\langle Sx, y\rangle| \le \|Sx\|\|y\| \le \|S\|\|x\|\|y\| \le \|S\|$$
Now, for any $x \in X$  with $Sx \ne 0$ we have:
$$\left|\left\langle Sx, \frac{Sx}{\|Sx\|}\right\rangle\right| = \frac{\|Sx\|^2}{\|Sx\|} = \|Sx\|$$
So taking the supremum over $\|x\| \le 1$ gives:
$$\sup_{\|x\|, \|y\| \le 1}|\langle Sx, y\rangle| \ge \sup_{\|x\| \le 1} \left|\left\langle Sx, \frac{Sx}{\|Sx\|}\right\rangle\right| = \sup_{\|x\| \le 1} \|Sx\| = \|S\|$$
We conclude $\displaystyle \sup_{\|x\|, \|y\| \le 1}|\langle Sx, y\rangle| = \|S\|$.
