Prove this surface integral implication

Suppose $$\iint\mathbf{F}\cdot d\mathbf{a}=0$$ for any closed surface. Prove that $$\iint\mathbf{F}\cdot d\mathbf{a}$$ is independent of surface, for any given boundary line.

My attempt:

I have the right idea but I'm having trouble with the signs.

Let $$A_1$$ and $$A_2$$ be two surfaces with the same boundary line.

$$A_1\cup A_2$$ is a closed surface. Or should it be $$A_1\cup(-A_2)$$?

The problem is that the sign of a surface integral is intrinsically ambiguous.

if $$\unicode{x222F} F\ dS = 0$$ for all closed surfaces, then the divergence of $$F =\nabla \cdot F= 0$$
If F is a divergence free field then there exists some $$G$$ such that the curl of $$G =\nabla \times G = F$$
$$\oint G\cdot dr = \iint \nabla \times G \ dA = \iint F \ dA$$