Norm closed convex iff weakly closed

Since weak-open sets are norm open, weak-closed sets are norm closed, and ${\bar{C}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}\subseteq {\bar{C}}^w$. If $C=\mathrm{\varnothing }$, both closures are empty.

For nonempty $C$, put $K={\bar{C}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}$. It is convex: for $u,v\in K$ and $0<t<1$, approximate $u,v$ by points $a,b\in C$ within any positive $\delta$; then $ta+(1-t)b\in C$ and its distance from $tu+(1-t)v$ is less than $\delta$. The cases $t=0,1$ are just $v,u\in K$. Thus $K$ is nonempty, closed and convex. For each $x\mathrm{\notin }K$, separation gives $f\in X^{\ast }$ and a real level $a$ with $\mathrm{Re}f(z)<a<\mathrm{Re}f(x)$ for all $z\in K$.

The set $\{y:\mathrm{Re}f(y)>a\}$ is weakly open, contains $x$ and misses $C$. Therefore $x\mathrm{\notin }{\bar{C}}^w$, giving ${\bar{C}}^w\subseteq K$ and equality of closures. If $C$ is norm closed this equality makes it weakly closed; the reverse implication was step 1.1. $\mathrm{\square }$