Problem
Assume $x_n \to 0$ weakly in a Banach space. Show that for all $\epsilon>0$ and for all $N\in \mathbb{N}$ there exists a $n>N$ s.t. for all $f\in X^\ast, \|f\|\leq 1$ there exists $N<m<n$ s.t. $|f(x_m)|\leq \epsilon$.
Attempt
Assume the statement doesn't hold, i.e. there is $\epsilon>0, N\in\mathbb{N}$ s.t. $\forall n>N$ there exists a $f_n\in X^\ast, \|f_n\|\leq 1$ with $|f_n(x_m)|\geq \epsilon$ for all $N<m<n$. Set $f$ as the a weak-*-limit point of $(f_n)_n$, which I can do due to Banach-Alaoglu. (I might eventually pass to a subsequence here.)
Fix $m>N$.Then forall $n>m$, I have $\epsilon\leq|f_n(x_m)|$. Then in particular $\epsilon\leq \lim_{n\to\infty}f_n(x_m) = f(x_m)$, contradicting the fact that $x_n\to 0$ weakly.
I am not sure about my proof and think that it is probably wrong, since I didn't use the notion of completeness of $X$.