I am dealing with 
exercise from Brezis.
I proved the existence of that $u_{n_k}$, and I have to prove that its weak derivative converges weakly in $L^p$, that is
$$\int_0^1 v(u'_{n_k}-u') \to 0, \ \ \ \forall v\in L^{p'}.$$
I tried to consider a sequence $(v_m)\subset C_c^{\infty}$, $v_m\to v$ in $L^{p'}$ for which I have (for $m$ fixed)
$$\int_0^1 v_m(u'_{n_k}-u')=-\int_0^1 v'_m(u_{n_k}-u) \to_{k\to\infty} 0$$
since $v'_m$ is bounded and $||u_{n_k}-u||_{L^{\infty}}\to 0$.
Now I would like to conclude taking the limit for $m\to \infty$, but I don't know if I can.
By hypothesis, $u_n'$ is bounded in $L^p$. By Banach-Alaoglu, we can find $v\in L^p$ and a subsequence (also written $u_n$) such that $u_n' \overset{*}{\rightharpoonup} v $. Notice that $v= u'$ since these are both distributional derivatives of $u$, and distributional derivatives are unique. In the case $p\in(1,\infty)$, $L^p$ is reflexive. Hence also $u'_n\rightharpoonup u'$ weakly in $L^p$.