Let $\Omega\subset\mathbb{R}^N$ be a open set, $\mathcal{D}(\Omega)=C_0^\infty(\Omega)$ and $D'(\Omega)$ the set of distributions. Suppose that $f_i\in \mathcal{D}'(\Omega)$ for $i=1,\ldots,N$. Define $f=(f_1,\ldots,f_N)$ and $$S=\{v\in (C_0^\infty(\Omega))^N:\ \operatorname{div}v=0\}$$
Assume that $\langle f,v\rangle=0$ for any $v\in S$, where $\langle f,v\rangle=\sum_{i=1}^N\langle f_i,v_i\rangle$. How can one prove that there exist $u\in \mathcal{D}'(\Omega)$ such that $\nabla u=f$?
Thank you.
This is a version of de Rham's result in his book Differentiable Manifolds ($\S 20$ to $\S 22$).
Luc Tartar has a presentation of a similar version of your result here, please see page 30 bottom, where $f\in \big(H^1_0(\Omega)\big)'\subset \mathcal{D}'$.
Amrouche and Girault extended the result to $f\in \big(W^{k,p}_0(\Omega)\big)'$ (Theorem 2.8 here) for Lipschitz $\Omega$, they said de Rham's theorem took too long to establish.
Both above literatures exploit close range theorem to prove, i.e., $\mathrm{Im}(\nabla) = \mathrm{ker}(\nabla\cdot)^{\perp}$.
This paper presented a special case of de Rham's theorem (please see Theorem 2.1 and 4.1), which is exactly your question.
I don't know if there is any proof based solely on functional analysis.