Let $V$ be a Hilbert space which is separable. Let $u_n \in L^2(0,T;V)$ with $u_n(t,x) = \sum_{i=1}^n u_{in}(t)w_i(x)$ where $u_{in}$ are absolutely continuous on $(0,T)$ and $w_i$ are a smooth basis of $V$. We have $u_n'$ exists.
Suppose that $u_n \rightharpoonup u$ in $L^2(0,T;V)$ where $u' \in L^2(0,T;V^*)$ exists and that $$\lim_{n \to \infty}\int_0^T \langle u_n', v \rangle = \int_0^T \langle u', v \rangle\quad\forall v \in C_c^\infty((0,T);V).$$
I wish to conclude that $u_n' \rightharpoonup u'$ in $L^2(0,T;V^*)$ (at least for a subsequence).
To do this, let us approximate $w \in L^2(0,T;V)$ by $w_j \in C_c^\infty(0,T;V)$ so $w_j \to w$ in $L^2(0,T;V)$. We have $$\lim_{n \to \infty}\int_0^T \langle u_n', w_j \rangle = \int_0^T \langle u', w_j \rangle.$$ Take limits over $j$: $$\lim_{j \to \infty}\lim_{n \to \infty}\int_0^T \langle u_n', w_j \rangle = \int_0^T \langle u', w \rangle.$$ So my question is, am I allowed to interchange the limits on the LHS so that I get what I want: $$\lim_{j \to \infty}\lim_{n \to \infty}\int_0^T \langle u_n', w_j \rangle = \lim_{n \to \infty}\lim_{j \to \infty}\int_0^T \langle u_n', w_j \rangle =\lim_{n \to \infty}\int_0^T \langle u_n', w \rangle ?$$
This is only possible if you know that $(u_n')$ is uniformly bounded in $L^2(0,T;V^*)$. Then write $$ \left|\int_0^T \langle u_n'-u',w\rangle \right| \le \left|\int_0^T \langle u_n'-u',w-w_j\rangle \right|+\left|\int_0^T \langle u_n'-u',w_j\rangle \right|\\ \le \|u_n'-u'\|_{L^2(0,T;V^*)}\|w-w_j\|_{L^2(0,T;V)}+\left|\int_0^T \langle u_n'-u',w_j\rangle \right|\\ $$ Take $\epsilon>0$. Since $(u_n'-u')$ is uniformly bounded in $L^2(0,T;V^*)$, you can choose one $j$ such that the first term is smaller than $\epsilon/2$ for all $n$. Now $j$ is fixed, and by your assumption the second integral is smaller than $\epsilon/2$ for all $n$ large enough.
Without uniform boundedness of $(u_n')$ the argument does not work. See the example in the answer there:
Weak convergence on a dense subset of test functions