= Solution
The <Frobenius theorem> says that <hypersurface orthogonality> of the nonzero covector $U_a$ is equivalent to $U_{[a}\nabla_bU_{c]}=0$, or locally $U_a=h\nabla_aS$ for some nonzero $h$. This immediately makes the screen projection of $\nabla_{[b}U_{a]}$ zero, since its antisymmetric part is a wedge product with $\nabla S$, which is proportional to $U$. Thus hypersurface orthogonality gives zero <null twist>.
For the converse, put $B_{ab}=\nabla_bU_a$. Nullness gives $U^aB_{ab}=\tfrac12\nabla_b(U^2)=0$, and the affine geodesic equation gives $B_{ab}U^b=0$. Resolve both tensor slots in the basis consisting of $U,N$ and the screen. These two identities exclude covector factors of $N$. Consequently
$$
B_{[ab]}=\widehat\omega_{ab}+U_{[a}q_{b]}
$$
for some covector $q$. If $\widehat\omega=0$, wedging this identity with $U$ makes $U_{[a}\nabla_bU_{c]}=0$. Applying <Frobenius theorem> completes the proof:
$$
\boxed{\widehat\omega_{ab}=0\iff\text{the null geodesic congruence is hypersurface-orthogonal}.}
$$
This is the <null twist vanishes exactly for hypersurface-orthogonal geodesics> criterion; the geodesic assumption ensures that the screen criterion implies the full Frobenius condition. The resulting hypersurfaces are null, and the congruence follows their null generators.
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