Every weakly convergent sequence is bounded Theorem: Every weakly convergent sequence in X is bounded.
Let $\{x_n\}$ be a weakly convergent sequence in X. Let $T_n \in X^{**}$ be defined by $T_n(\ell) = \ell(x_n)$ for all $\ell \in X^*$. Fix an $\ell \in X^*$. For any $n \in \mathbb{N}$, since the sequence $\{\ell(x_n)\}$ is convergent, $\{T_n(\ell)\}$ is a bounded set. By Uniform Boundedness Principle $ \sup_{n \in \mathbb{N}} \|x_n\| = \sup_{n \in \mathbb{N}} \|T_n\| < \infty,$ i.e. $\{x_n\}$ is bounded.
My question is: why $ \sup_{n \in \mathbb{N}} \|x_n\| = \sup_{n \in \mathbb{N}} \|T_n\|$ ?
 A: The equality $\|x_n\|=\|T_n\|$ is an instance of the fact that the canonical embedding into the second dual is an isometry.
See also Weak convergence implies uniform boundedness which is stated for $L^p$ but the proof works for all Banach spaces.
A: One way to think about it is to recall the following 'duality property' for norms, which holds for any normed space $(X,\lVert{\cdot}\rVert)$:
$$ \forall x \in X: \lVert x \rVert = \sup_{\lVert f\rVert_*=1} \lvert f(x) \rvert $$
$$ \forall f \in X^*: \lVert f \rVert_* = \sup_{\lVert x\rVert =1} \lvert f(x) \rvert $$
where the second equality is almost a tautology, depending on your definition of the dual norm $\lVert \cdot \rVert_*$ and the first equality is a consequence of the Hahn-Banach Theorem and is a pretty easy exercise. From this, the fact that $\lVert T_x \rVert_{**} \equiv \lVert x \rVert$ is pretty obvious:
$$ \lVert T_x \rVert_{**} = \sup_{\Vert f \Vert_* = 1} \vert T_x(f) \vert = \sup_{\Vert f \Vert_* = 1} \vert x(f) \vert = \lVert x \rVert$$
