Let $X$ be a Banach space and let $x_n$ be a sequence that converges to zero in the weak topology. Assuming that the set $\{x_1, x_2, x_3, \dots \}$ is totally bounded, does the sequence $(x_n)$ converges to zero in norm?
The fact that $x_n$ converges to zero in the weak topology implies that if $(x_n)$ converges in $X$, then it must converges to zero. So it is sufficient to show that $(x_n)$ is a Cauchy sequence. If not, exists an $\epsilon > 0$ such that for all $n$, exists $l_n, m_n \geq n$ such that $$ \| x_{l_n} - x_{m_n} \| < \epsilon. $$ By the other side, exists $n_{1}, \dots, n_{k}$ such that $$ \{x_n\}_n \subset \bigcup_{j=1}^k B(x_{n_j}, \epsilon). $$
I'm not so sure how to get a contradiction here.
Help?
It suffices to show that any subsequence of $\{x_n\}_n$ has a subsequence that converges to $0$ in norm, so fix a subsequence $\{x_{n_k}\}_k$ of $\{x_n\}_n$. $\{x_{n_k}\}_k$ lies in the set $\overline{\text{span}\{x_n :n\in \mathbb{N}\}}$ which is norm-compact because it is totally bounded and closed, hence $\{x_{n_k}\}_k$ has a subsequence, say $\{x_{n_{k_c}}\}_c$ which converges to an $x\in X$ in norm. Since $\{x_{n_{k_c}}\}_c$ is known to converge weakly to $0$ we must have $x=0$ so we are done.