Weak convergence in Sobolev space $H_1$ implies weak convergence in $L^2$ I am trying to prove this statement

Let $D \subset \mathbb{R}^m$. If $\{v_n\}$ is a weakly converging sequence in Sobolev spapce $H_1(D)$, then $\{v_n\}$ weakly converges in $L^2(D)$.

This is the exercise 10.12 from the book An Introduction to Partial Differential Equations by Y.Pinchover and J. Rubinstein. I followed the solution manual.
This is my attempt:
Denote the weak limit of $\{v_n\}$ with $v$. Because $\{v_n\}$ is a weakly convergent sequence, that means that it is bounded so there exists a $M \geq 0$ so that
\begin{align*}
    &||v_n||_{H_1(D)} \leq M  \quad \text{for every } n = 1, 2 \dots \\
\implies &||v_n||^2_{H_1(D)} = \int_D ((\nabla v_n)^2 + v_n^2) d{\vec{x}} \leq M^2 \quad \text{for every } n = 1, 2 \dots\\
    \implies &\int_D (\nabla v_n)^2 d{\vec{x}} = ||\nabla v_n ||_{L^2(D)}^2 \leq M^2 \\
    \implies &\int_D v_n^2 d{\vec{x}} = || v_n||_{L^2(D)}^2 \leq M^2.
\end{align*}
That means that $\{v_n\}$ and $\{\nabla v_n\}$ are also bounded in $L^2(D)$.
Since the sequence $\{\nabla v_n\} = \{(\frac{\partial v_n}{\partial x_1}, \frac{\partial v_n}{\partial x_2}, \dots, \frac{\partial v_n}{\partial x_m})\}_{n=1}^{\infty}$ is bounded in $L^2(D)$, the sequences $\{\frac{\partial v_n}{\partial x_i}\}_{n=1}^{\infty}$ are also bounded in $L^2(D)$ for every $i = 1, 2, \dots, m$. This means that the sequences $\{v_n\}$ and $\{\frac{\partial v_n}{\partial x_i}\}$ contain a weakly convergent subsequence in $L^2(D)$.
Denote the weak limit of $\{v_{n_k}\}_k$ with $\tilde{v}$ and the weak limit of $\{\frac{\partial v_{n_k}}{\partial x_i}\}$ with $\tilde{v_i}$ for every $i = 1, 2, \dots, m$. This means that for every $u \in {L^2(D)}$ and $i = 1, 2, \dots, m$
$ \langle \tilde{v}_i, u\rangle_{L^2(D)} =\lim\limits_{k \to \infty}  \langle \frac{\partial v_{n_k}}{\partial x_i}, u\rangle_{L^2(D)}$.
If we take $u$, such that $u_{|\partial D} = 0$ then
\begin{align*}
    \langle \tilde{v}_i, u\rangle _{L^2(D)}
    &= \lim\limits_{k \to \infty} \int\limits_D \frac{\partial v_{n_k}}{\partial x_i} u d{\vec{x}}\\
    &= -\lim\limits_{k \to \infty} v_{n_k} \frac{\partial u}{\partial x_i} d{x} \\
    &= - \int\limits_D \tilde{v} \frac{\partial u}{\partial x_i} d{x} \\
    &= \int\limits_D \frac{\partial \tilde{v}}{\partial x_i} u d{x} \\
    & = \langle \frac{\partial \tilde{v}}{\partial x_i} ,u\rangle_{L^2(D)}.
\end{align*}
This means $\tilde{v_i} = \frac{\partial \tilde{v}}{\partial x_i}$ for every $i = 1, 2, \dots, m$.
I do not quite know what to do with this. - I think I am missing the final conclusion, I also do not understand why I even needed to show $\tilde{v_i} = \frac{\partial \tilde{v}}{\partial x_i}$. Did I also prove that $\{\nabla v_n\}$ is weakly convergent in $L^2(D)$?
 A: It is unnecessary to make things this knid complicated. Anyway, I'll finish the proof you presented in OP, and then I'll give a more elegant proof.
$\tilde{v_i} = \frac{\partial \tilde{v}}{\partial x_i}$ implies that $\frac{\partial \tilde{v}}{\partial x_i}\in L^2$, thus $\tilde{v}\in H_1$ and then we know that $v_{n_k}\rightharpoonup\tilde v$ in $H_1$. Since we also have $v_n\rightharpoonup v$ in $H_1$, it follows that $\tilde v=v$ a.e.
Now, we have proved that
$$\text{if}\ \ v_n\rightharpoonup v \ \ \text{in}\ \ H_1, \text{then there exists a subsequence} \ \ \{v_{n_k}\}\ \  \text{such that }\  v_{n_k}\rightharpoonup v \ \text{ in}\ \ L^2(D).$$
This will imply that $v_n\rightharpoonup v$ in $L^2$. Indeed, fix any $u\in L^2(D)$, we need to show that
$$\lim_{n\to \infty}\langle v_n,u\rangle_{L^2}=\langle v,u\rangle_{L^2}.\tag{1}$$
We use the fact that $\lim_{n\to\infty}a_n=A$ if and only if every subsequence of $\{a_n\}$ has a sub-subsequence converging to $A$, see here for the proof of this result. Now, for a subsequence $\langle v_{n_k},u\rangle_{L^2}$ of $\langle v_n,u\rangle_{L^2}$, since $v_{n_k}\rightharpoonup v$ in $H_1$, by the third paragraph, there is a sub-subsequence $v_{n_{k_j}}$ of $v_{n_k}$ converging weakly to $v$ in $L^2$, thus
$$\lim_{j\to \infty}\langle v_{n_{k_j}},u\rangle_{L^2}=\langle v,u\rangle_{L^2}.$$
Now, $(1)$ follows and the proof is complete.
About $\{\nabla v_n\}$: the same trick implies that $\nabla v_n\rightharpoonup\nabla v$ in $L^2$.
Another method. For each $u\in L^2$, the linear functional
$$T_u: H_1\to\mathbb R, w\mapsto \langle w,u\rangle_{L^2}$$
is continuous since $|T_u(w)|\leq \|w\|_{L^2}\|u\|_{L^2}\leq \|u\|_{L^2}\|w\|_{H_1}$ for all $w\in H_1$. Hence $T_u(v_n)\to T_u(v)$, i.e.
$$\lim_{n\to \infty}\langle v_n,u\rangle_{L^2}=\langle v,u\rangle_{L^2}.$$
This proves the weak convergence of $\{v_n\}$ in $L^2(D)$.
