Let $ |X_1| = m$, $ |X_2| = n$ and $ |X| = \ell$. ($ \ell = m + n$) For each pair $ \{a,b\}$ from $ X_1$ we have $ p$ different subsets $ A_i$ Then there are in total $ {m\choose 2}\cdot p$ different subsets $ A_i$ (beacuse each $ A_i$ cuts both $ X_1$ and $ X_2$) Also there is $ {n\choose 2}\cdot p$ (beacuse of $ X_2$) So we have at least $ {m\choose 2}\cdot p + {n\choose 2}\cdot p$ subsets $ A_i$: $ k \geq {m\choose 2}\cdot p + {n\choose 2}\cdot p \geq {\ell ^2 - \ell\over 4}\cdot p$ On the other side each pair $ \{a,b\}$ from $ X$ is ''connected'' with $ p$ diffrent subsets $ A_i$ and each subset $ A_i$ is connected with exactly $ 3$ pairs $ \{a,b\}$, so we have \[ {\ell \choose 2}\cdot p = 3k \] Joining both equations we get $ \ell\leq 4$.