= Solution
By <weakly convergent sequence is bounded>, $M=\sup_n\|x_n\|^2$ is finite. Choose any $n_1$. Once $n_1<\cdots<n_k$ have been chosen, put $s_k=\sum_{j=1}^kx_{n_j}$. Since $s_k$ is fixed, <weak convergence> to zero gives $\langle s_k,x_n\rangle\to0$. Choose $n_{k+1}>n_k$ with $|\langle s_k,x_{n_{k+1}}\rangle|\leq1$. This gives the required one-sided inequality as well.
The useful estimate is for the \b[squared] <norm>. Indeed, $\|s_1\|^2\leq M$, and the <inner product> expansion gives
$$
\|s_{k+1}\|^2
=\|s_k\|^2+2\langle s_k,x_{n_{k+1}}\rangle+\|x_{n_{k+1}}\|^2
\leq\|s_k\|^2+2+M.
$$
By <mathematical induction>, $\|s_k\|^2\leq Mk+2(k-1)\leq(M+2)k$. Hence
$$
\boxed{\left\|\frac1m\sum_{k=1}^m x_{n_k}\right\|^2\leq\frac{M+2}{m}\longrightarrow0.}
$$
This proves <norm convergence> of the averages, the <weak Banach–Saks theorem in a Hilbert space>. The printed inequality omits the square on its left side. Our stronger estimate also proves that printed bound, since $\sqrt{(M+2)k}\leq(M+2)k$ for $k\geq1$, but the printed bound alone only bounds the averages and cannot imply their <norm convergence> to zero.
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