Example of normed linear space with bounded sequence in dual having no weak* convergent subsequence

If $X$ is a separable normed linear space, then we know that every bounded sequence in $X^*$ has a weak-* convergent subsequence . Can we drop the separability condition , i.e. if we don't assume $X$ is separable, then are there counterexamples ?

• @David Mitra : What is meant by $l_1(\mathbb R)$ on the book ? Isn't it kind of contradictory as I thought $l_1$ is separable ... – user495643 Nov 17 '17 at 11:44
• The space of absolutely convergent sums, indexed by $\Bbb R$; so elements are of the form $\sum_{r\in\Bbb R} x_r$. – David Mitra Nov 17 '17 at 11:46
Take $X=l^\infty$, which is not separable. Define the sequence $f_n$ in $(l^\infty)^*$ by $$f_n (x) = x_n.$$ Then $(f_n)$ is a bounded sequence, in fact, $\|f_n\|_{(l^\infty)^*}=1$.
However, it does not have a weak-star converging subsequence. Let $(f_{n_k})$ denote a subsequence. Then define $x\in l^\infty$ by $$x_{n_k}=(-1)^k,$$ set all other entries $x_i=0$. Then $$f_{n_k}(x) = (-1)^k,$$ which is not convergent.