= Solution
The <weak topology> is coarser than the <norm> topology, so <norm convergence> implies <weak convergence>, and
$$
\overline C^{\|\cdot\|}\subseteq\overline C^{\mathrm{weak}}.
$$
For the reverse inclusion, let $x\notin K=\overline C^{\|\cdot\|}$. The set $K$ is closed and convex. The <Hahn-Banach separation theorem> supplies $f\in X^*$ and $a\in\mathbb R$ such that $f(x)>a\geq\sup_{y\in K}f(y)$. The set $\{z:f(z)>a\}$ is weakly open, contains $x$ and misses $C$. Hence $x$ is outside the weak <closure> as well. If $C$ is empty the equality is immediate. Therefore
$$
\boxed{\overline C^{\mathrm{weak}}=\overline C^{\|\cdot\|}\quad\text{for every convex }C.}
$$
The separating functional explains the role of convexity: an arbitrary norm-closed subset need not be weakly closed.
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