I think some care needs be taken at the point, where you write
[...] The LHS converges to $\int_0^T v(t) \phi(t)dt$ and the RHS to $-\int_0^T u(t) \phi'(t)dt$. Therefore [...]
This is true, but actually not given in the exercise statement, and I believe its deduction is the main part of the problem. Otherwise you'd already be done, having shown that $\int_0^T u(t)\phi'(t)\, dt = -\int_0^T v(t)\phi(t)\, dt$ for all $\phi\in C_c^1(0,T)$. A function $v$ (by definition) is a weak derivative of $u$ if this holds.
Maybe we should first recall what it means for a sequence $w_n$ to converge to $w$ weakly in $L^2(0,T; X^\ast)$, where $X$ is some Banach space. By definition (at least if I'm not misunderstanding something myself) this means that
$$\int_0^T \langle w_n, g\rangle \, dt \to \int_0^T \langle w,g\rangle\, dt \qquad \forall \, g \in L^2(0,T; X)$$
where $\langle\; , \;\rangle $ is the natural pairing $X^\ast \times X \to \mathbb R$.
Now following the hint: If we set $X = H^1_0(U)$, $X^\ast= H^{-1}(U)$ and $g = \phi(t) w$ for $\phi\in C_c^1(0,T)$, $w\in H^1_0(U)$, then $g, g'\in L^2(0,T;H^1_0(U))$. So by assumption on $u$ and $v$:
\begin{align}
\int_0^T \langle u_n, \phi' w\rangle \, dt &\to \int_0^T \langle u,\phi' w\rangle\, dt \\
\int_0^T \langle u_n', \phi w\rangle \, dt &\to \int_0^T \langle v,\phi w\rangle\, dt
\end{align}
We can rewrite
$$\int_0^T \langle u, \phi' w\rangle \, dt = \int_0^T \phi' \langle u, w\rangle \, dt = \int_0^T \langle u\phi' , w\rangle \, dt = \left\langle \left(\int_0^T u(t)\phi'(t) \, dt\right) , w\right \rangle$$
where I have made use of Fubini in the last equality. A similar equality is true for $v$. Using this together with $\int_0^T \langle u_n, \phi' w\rangle \, dt = -\int_0^T \langle u_n', \phi w\rangle \, dt$, we obtain
$$\left\langle \left(\int_0^T u(t)\phi'(t) \, dt\right) , w\right \rangle = \left\langle \left(\int_0^T -v(t)\phi(t) \, dt\right) , w\right \rangle \qquad \forall\, w\in H^1_0(U)$$
Therefore $\int_0^T u(t)\phi'(t) \, dt = -\int_0^T v(t)\phi(t) \, dt$ for all $\phi \in C_c^1(0,T)$. So $v$ is a weak derivative of $u$.
Best Answer
Let $I\subset\Bbb{R}$ be an open interval, $U\subset\Bbb{R}^n$ an open set, and $\Phi:I\times U\to\Bbb{R}^n$ a map with whatever smoothness you like: $C^1,C^k,C^{\infty},C^{k,\alpha}$, mixed regularity in the two variables etc. Let us adopt the notation $\Phi_{t}(x);\Phi(t,x)$. Then, we call the family of sets $U(\tau):=\Phi_{\tau}[U]$ a smooth 1-parameter family of sets. Now, typically, one imposes the condition that for each $\tau\in I$, the set $\Phi_{\tau}[U]$ is open in $\Bbb{R}^n$ and the map $\Phi_{\tau}:U\to\Phi_{\tau}[U]\subset\Bbb{R}^n$ is a diffeomorphism onto its image (so it can be considered as a “time-dependent change of variables”).
You can of course generalize this: given three smooth manifolds $X,Y,Z$ and a smooth map $\Phi:X\times Y\to Z$, you can talk about a smooth parameter-dependent family of sets, namely $\Phi_x[Y]$ as $x$ varies in $X$ (so you think of $X$ as the space of paramaeters, and think of $\Phi_x[Y]$ as a copy of $Y$ sitting inside $Z$ (assuming $\Phi_x$ is a diffeomorphism onto its image)).