I am working with a space of the form $E=L^{\infty}(0,\infty;X)$, where $X$ is a non reflexive Banach space.
If a sequence $(x_n)$ is bounded in $E$, how to extract a weakly converging subsequence?
I think that this is not always possible for an arbitrary Banach space $X$!
I tried to consider that $X$ has a separable dual and use Banach Alaoglu theorem, but I did not success to identify the dual of $E$.
What I am looking is to assume that $X$ satisfies a given property letting me extract a weakly converging subsequence.
Any help will be appreciated.
As you thought, this is indeed not always possible.
Since $X$ isn't reflexive, a classical consequence of the Mackey-Arens theorem gives you the existence of bounded subsets of $X$ which aren't weakly relatively compact in $X$ and the Eberlein-Šmulian theorem then gives you the existence of a bounded sequence $(y_n)_{n\in\mathbf N}$ of elements of $X$ which has no weakly convergent subsequence.
For every $n\in\mathbf N$, let $x_n$ be (the equivalence class in $L^\infty(0,\infty;X)$ of ) the constant function on $(0,\infty)$ taking everywhere the value $y_n$. The sequence $(x_n)_{n\in\mathbf N}$ is a bounded sequence of $L^\infty(0,\infty;X)$ and I will show that it has no weakly convergent subsequence.
Let's proceed by contradiction and suppose that $(x_n)_{n\in\mathbf N}$ has a subsequence $(x_{n_k})_{k\in\mathbf N}$ which converges weakly to some $x_\infty\in L^\infty(0,\infty;X)$. By the Lebesgue differentiation theorem, $x_\infty$ has many Lebesgue points (maybe you only know this theorem for scalar functions, but the proof presented on the Wikipedia page can be adapted to the case of vector-valued functions without any major modification); let $t\in(0,\infty)$ be a Lebesgue point of $x_\infty$ and let $$y_\infty=\lim_{h\to 0^+}\frac 1{2h}\int_{t-h}^{t+h}x_\infty(s)\,ds.$$ I will show that $(y_{n_k})_{k\in\mathbf N}$ converges weakly to $y_\infty$, which will be a contradiction because $(y_n)_{n\in\mathbf N}$ has no weakly convergent subsequence.
Let $g$ be a continuous linear functional on $X$. Let $\mathscr U$ be an ultrafilter on $(0,t)$ such that $(0,h)\in\mathscr U$ for every $h\in(0,t)$. The linear functional $$f:x\longmapsto\lim_{h,\mathscr U}\frac 1{2h}\int_{t-h}^{t+h}g(x(s))\,ds$$ on $L^\infty(0,\infty;X)$ can be easily checked to be well-defined and continuous. Moreover, because $(x_{n_k})_{k\in\mathbf N}$ converges weakly to $x_\infty$ and $g(y_n)=f(x_n)$ for every $n\in\mathbf N\cup\{\infty\}$, we have $$g(y_{n_k})=f(x_{n_k})\underset{k\to\infty}{\longrightarrow}f(x_\infty)=g(y_\infty),$$ which concludes.