You do not know that $\lim\limits_{n\rightarrow\infty} \Vert x_n\Vert$ exists. For instance, the sequence $(e_1, 2e_2, e_3, 2e_4,\ldots)$ converges weakly to $0$ in $\ell_2$.
But, as the $\Vert x_n\Vert$ are reals, $\liminf\limits_{n\rightarrow\infty}\Vert x_n\Vert$ exists, and you can find a subsequence $\Vert x_{n_k}\Vert$ converging to its value. Then since $(x_{n_k})$ converges weakly to $x$ $\tag{1} |f(x)|= \lim\limits_{k\rightarrow\infty}|f(x_{n_k})| \le \lim\limits_{k\rightarrow\infty}(\,\Vert x\Vert\Vert x_{n_k}\Vert\,) =\Vert x\Vert \liminf\limits_{n\rightarrow\infty}\Vert x_{n }\Vert. $
Here we are just using the result for real numbers: Suppose $a_n\le b_n$ for each $n$. Then if $a_n\rightarrow a$ and if $b_n\rightarrow b$, it follows that $a\le b$.