= Solution
Put $z_n=x_n-x$, so $z_n\rightharpoonup0$. We show that the norm closure of the <convex hull> of every tail
$$
A_k=\{z_n:n\geq k\}
$$
contains zero. If it did not, the <Hahn-Banach separation theorem> would give a bounded linear functional $\phi$, a real number $c>0$, and, after multiplying $\phi$ by a complex scalar if necessary,
$$
\operatorname{Re}\phi(z)\geq c
$$
for every $z$ in that closed convex hull. In particular $\operatorname{Re}\phi(z_n)\geq c$ for all $n\geq k$. By the <Riesz representation theorem>, $\phi(z)=\langle z,y\rangle$ for some $y\in H$, contradicting $z_n\rightharpoonup0$.
Consequently, for every $k$ there is a finite <convex combination>
$$
w_k=\sum_{n=k}^{N_k}\lambda_{k,n}z_n,
\qquad
\lambda_{k,n}\geq0,
\qquad
\sum_{n=k}^{N_k}\lambda_{k,n}=1,
$$
with $\lVert w_k\rVert_H<1/k$. Define
$$
\widetilde x_k=x+w_k
=\sum_{n=k}^{N_k}\lambda_{k,n}x_n.
$$
Then every $\widetilde x_k$ is a convex combination of terms of the original sequence and
$$
\lVert\widetilde x_k-x\rVert_H=\lVert w_k\rVert_H<\frac1k\longrightarrow0.
$$
This is <Mazur lemma> in the present Hilbert-space setting.
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