= Solution
The <Uniform boundedness principle> states that if $E$ is a <Banach space> and a family of <bounded linear operators> $S_\alpha:E\to Y$ is pointwise bounded, then $\sup_\alpha\lVert S_\alpha\rVert<\infty$.
Suppose $x_n\rightharpoonup x$ in the <normed vector space> $X$. For each $n$, define the evaluation functional
$$
J_n:X^*\longrightarrow\mathbb K,
\qquad J_n(f)=f(x_n).
$$
The <completeness of the dual space> says that $X^*$ is Banach even when $X$ is not. For each $f\in X^*$, the scalar sequence $J_n(f)=f(x_n)$ converges to $f(x)$ and is therefore bounded. Uniform boundedness gives
$$
\sup_n\lVert J_n\rVert<\infty.
$$
By the <canonical embedding into the bidual> and the <Hahn-Banach theorem>, $\lVert J_n\rVert=\lVert x_n\rVert$. Hence
$$
\boxed{\sup_n\lVert x_n\rVert<\infty,}
$$
which proves that a <weakly convergent sequence is bounded>.
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