 Solution

First suppose the convex set $K$ is closed in the norm topology, and $x_n\in K$ satisfies $x_n\rightharpoonup x$. Apply the weak Banach–Saks theorem in a Hilbert space to the weakly null sequence $x_n-x$. Along the resulting subsequence,in norm. Each $z_m$ is a convex combination of points in $K$, so $z_m\in K$. Since $K$ is a closed set in the norm topology, $x\in K$.

Conversely, suppose $K$ is a sequentially weakly closed set. If $x\in \overline{K}$ in the norm topology, choose $x_n\in K$ with $\Vert x_n-x\Vert <1/n$. The Cauchy-Schwarz inequality gives $|⟨x_n-x,z⟩|\le \Vert x_n-x\Vert \Vert z\Vert \to 0$ for every $z$, hence weak convergence to $x$. The assumed closure property gives $x\in K$. The empty set satisfies both properties too. Thus