prerequisite: For $\underline{q} \in H(\operatorname{div}, \Omega)$,we can define $\left.\underline{q} \cdot \underline{n}\right|_{\Gamma} \in H^{-\frac{1}{2}}(\Gamma)$ and $$ \int_{\Gamma} \underline{q} \cdot \underline{n} v d \sigma=\int_{\Omega} \operatorname{div} \underline{q} v d x+\int_{\Omega} \underline{q} \cdot \operatorname{grad} v d x \ \ \ \ \ \forall v\in H^{1}(\Omega) $$
Then in the textbook, it says :
Lemma 2.1.2. The trace operator $\left.\underline{q} \in H(\operatorname{div} ; \Omega) \rightarrow \underline{q} \cdot \underline{n}\right|_{\Gamma} \in H^{-\frac{1}{2}}(\Gamma)$ is surjective.
Proof. Let $g \in H^{-\frac{1}{2}}(\Gamma)$ be given. Then, solving in $H^{1}(\Omega)$ $$ \int_{\Omega} \operatorname{grad} \phi \cdot \operatorname{grad} v d x+\int_{\Omega} \phi v d x=\langle g, v\rangle, \forall v \in H^{1}(\Omega) $$ and making $\underline{q}=\operatorname{grad} \phi$ implies $\left.\underline{q} \cdot n\right|_{\Gamma}=g$.
I think the proof is wrong...
The first doubt is why solving in $H^{1}(\Omega)$ $$ \int_{\Omega} \operatorname{grad} \phi \cdot \operatorname{grad} v d x+\int_{\Omega} \phi v d x=\langle g, v\rangle, \forall v \in H^{1}(\Omega) $$ instead of solving $$ \int_{\Omega} \operatorname{div} \underline{q} v d x+\int_{\Omega} \underline{q} \cdot \operatorname{grad} v d x = <g,v>\ \ \ \ \ \forall v\in H^{1}(\Omega) $$ The second doubt is about the last sentence in his proof making $\underline{q}=\operatorname{grad} \phi$. Besides, we just have $\phi \in H^{1}$, why it is sufficient to illustrate $\operatorname{grad}\phi \in H(\operatorname{div},\Omega)$
First question: because the $H^1$-problem is solvable.
Second: Take a smooth test function $v\in C_c(\Omega)$. Then $\langle g,v\rangle=0$. This implies $$ \int_\Omega \nabla \phi \nabla v = -\int_\Omega \phi v $$ for all $v\in C_c(\Omega)$, which implies $div(\nabla\phi) =\phi\in L^2(\Omega)$ in the weak sense.