Let $X$ and $Y$ be Banach spaces and $B(\cdot, \cdot): X \times Y \rightarrow \mathbb{C}$ a separately continuous bilinear mapping. In proving that $B(\cdot, \cdot)$ is jointly continuous, let $T_n(y) = B(x_n, y)$ for some sequence $\{x_n\}$.
In their book, Reed & Simon state that if $x_n \rightarrow 0$, and since $B(\cdot, y)$ is bounded, $\{\|T_n(y)\|\}$ is bounded for each fixed $y$. Thus there exists $C$ so that $$\|T_n(y) \| \leq C\|y\| \quad \forall n.$$
The book does not mention it explicitly, but I am guessing they use the principle of uniform boundedness to get $C$ to be uniform in $n$. If so, why is there $\|y\|$ on the right hand side?
How does $x_n \rightarrow 0$ ensure that $\{\|T_n(y)\|\}$ is bounded for each fixed $y$?
Since $B$ is separately continuous and bilinear, you know that $\vert B(x,y)\vert \leq C_y \Vert x\Vert $. This gives you for every fixed $y$ that
$$ \vert T_n(y)\vert =\vert B(x_n,y)\vert \leq C_y \cdot \sup_n \Vert x_n\Vert. $$
Since $x_n\to 0$, you know that $\sup_n \Vert x_n\Vert<\infty$. The principle of uniform boundedness gives you that $\sup_{n}\Vert T_n\Vert \leq C$. Therefore, you know by linearity of the $T_n$'s that
$$ \vert T_n(y) \vert \leq \Vert T_n\Vert \cdot \Vert y\Vert \leq C\cdot \Vert y\Vert. $$