There is a simple proof along the following lines: Because the first deRham cohomology group of $\mathbb{RP}^2$ is trivial, a $1$-form on $\mathbb{RP}^2$ is exact if and only if it is closed. Let $\alpha$ be a $1$-form on $\mathbb{RP}^2$ whose integral over every line vanishes, and let $\beta = \pi^*\alpha$ where $\pi:S^2\to \mathbb{RP}^2$ is the standard double cover. Then the integral of $\beta$ over every great circle vanishes and $\beta$ is invariant under the antipodal involution $\iota:S^2\to S^2$, so $d\beta$ is a $2$-form on $S^2$ that is $\iota$-invariant and its integral over every hemisphere must vanish.
Now $d\beta = B\ dA$ where $dA$ is the standard volume form on $S^2$ and we must have $B\circ\iota=-B$ in order for $d\beta$ to be invariant. Now you use the representation theoretic fact that the operation $A:C^\infty(S^2)\to C^\infty(S^2)$ defined by $$ Af(u) = \frac1{2\pi}\int_{v\cdot u\ge0} f\ dA $$ is $\mathrm{SO}(3)$-equivariant and hence must, on each eigenspace of the Laplacian on $S^2$ be a multiple of the identity (since these eigenspaces are irreducible representations of $\mathrm{SO}(3)$). To complete the proof, you just need to check that $A$ is nonzero on the odd eigenspaces, and this should be straightforward (just evaluate on a rotationally symmetric candidate from each eigenspace).