Weakly compact implies bounded in norm The weak topology on a normed vector space $X$ is the weakest topology making every bounded linear functionals $x^*\in X^*$ continuous.
If a subset $C$ of $X$ is compact for the weak topology, then $C$ is bounded in norm.
How does one prove this fact? 
 A: The first key point is that an element of $x$ can be identified with a linear functional of norm $\|x\|$ on the dual $X^*$. Indeed, it follows from Hahn-Banach that there exists $x^*\in X^*$ such that $x^*(x)=\|x\|$ with $\|x^*\|=1$. Therefore, denoting by $e_x$ the linear functional (called point evaluation)  $e_x:x^*\longmapsto x^*(x)$ on $X^*$, we have
$$
\|e_x\|=\sup_{\|x^*\|\leq 1}|e_x(x^*)|=\sup_{\|x^*\|\leq 1}|x^*(x)|=\max_{\|x^*\|\leq 1}|x^*(x)|=\|x\|.
$$
This yields the canonical isometric embedding of $X$ into the double dual $X^{**}$.
The second key point is that $X^*$ is always a Banach space. Therefore we can use the uniform boundedness principle (Banach-Steinhaus) for bounded linear functionals on $X^*$, that is in $X^{**}$.
If $C$ is weakly compact in $X$, then the image of $C$ under every $x^*\in X^*$ is compact, hence bounded, in the base field. Just because $x^*$ is continuous for the weak topology by definition of the latter, and because the continuous image of a compact space is compact. It follows that
$$
\sup_{x \in C}|x^*(x)|=\sup_{x \in C}|e_x(x^*)|<\infty \qquad \forall x^*\in X^*.
$$
By the uniform boundedness principle applied to $\{e_x\,;\,x\in C\}$, this implies that
$$
\sup_{x\in C}\|x\|=\sup_{x\in C}\|e_x\|<\infty
$$
which says precisely that $C$ is bounded in norm.
Note: since the weak topology is Hausdorff, we get that weakly compact implies weakly closed + norm bounded. The converse is false. For instance, the closed unit ball of $c_0$ is weakly closed (the weak closure of a convex set is the same as its norm closure) and norm bounded, but not weakly compact (the closed unit ball of a normed vector space $X$ - actually automatically a Banach space in either case - is weakly compact if and only if $X$ is reflexive). Things are better with the weak*-topology on the dual of a Banach space $X$: weak*-compact is equivalent to weak*-closed + norm bounded.
