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).

*Added comment:*  Here's a simple way to see that $A$ is nonzero on the odd eigenspaces (without having to do any integration). The $d$-th eigenspace $H_d$ of the Laplacian on $S^2$ is simply the restriction to $S^2$ of the harmonic polynomials on $\mathbb{R}^3$ that are homogeneous of degree $d$.  Since $A$ acts as a multiple of the identity on each of these spaces, it's enough to show that, when $d=2m{+}1$, $A$ is nonzero on at least one element of $H_d$.  For this purpose, let $x$, $y$, and $z$ be the standard coordinates on $\mathbb{R}^3$ and consider the polynomial
$$
f = \mathrm{Re}\bigl((x+i y)^{2m+1}\bigr) 
= \prod_{k=-m}^m \left( x-\tan\left(\frac{k\pi}{2m{+}1}\right) y\right).
$$
Clearly $f$ belongs to $H_{2m+1}$ and $f(1,0,0)=1\not=0$.  Moreover, $f$ is odd with respect to rotation by an angle of $\pi/(2m{+}1)$ about the $z$-axis, and it vanishes on the $2m{+}1$ planes given by the above factors, which divide the sphere into $4m{+}2$ congruent 'sectors', on half of which $f$ is positive and on half of which $f$ is negative. Now, $Af(1,0,0)$ is the average of $f$ over the hemisphere where $x\ge0$, and this is composed of $2m{+}1$ of the sectors, which alternate in the sign of $f$.  Since there is an odd number of them, they can't cancel out, so $Af(1,0,0)$ cannot be zero.  Thus, $A$ is nonzero on the odd harmonics, as was to be shown.