This is sort of a hypergraph-ish question that I feel should be easy to prove or disprove but I can't see it right now.

The setup is as follows. We have a vertex set partitioned in to sets $V_1,\ldots,V_\ell$ of size 8, and transversing these are sets $\mathcal S = \{ S_1,\ldots, S_r \}$ of size at most 9.

Each $S_i$ intersects each $V_j$ at most once.

Each vertex is in at most two sets of $\mathcal S$.

We will partition each $V_i$ into two sets $A_i$ and $B_i$, each of size 4. Let $A$ denote $\cup A_i$ and let $B$ denote $\cup B_i$.

The question is, can we find $A$ and $B$ such that no $S_i$ has at least four vertices in $A$ and at least four vertices in $B$? That is, each $S_i$ has at most three in $A$ or at most three in $B$.

Even better would be $A$ and $B$ such that each $S_i$ has at most two in $A$ or at most two in $B$, but I find this hard to believe.

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