Yes, your conjecture is true.

Suppose otherwise. Then there exists a counterexample $f : \mathcal{P}(8) \rightarrow \{0, 1\}$. For each set $X \in \mathcal{P}(8)$, let the proposition $P_X$ denote $f(X) = 1$.

There are $5440$ different choices of the tuple $(A, B, C) \in \mathcal{P}(8)^3$ satisfying your constraints. For each such tuple, we obtain two clauses which must be true of the counterexample $f$:

$$ (\neg P_C \lor \neg P_{A \cup C} \lor \neg P_{B \cup C} \lor \neg P_{A \cup B \cup C}) $$

$$ (P_C \lor P_{A \cup C} \lor P_{B \cup C} \lor P_{A \cup B \cup C}) $$

This gives a succinct list of $10880$ clauses which must be true of the $256$ primitive propositions $\{ P_X : X \in \mathcal{P}(8) \}$.

Inputting this list of clauses into the SAT solver *Glucose* gives the response 'UNSAT' (meaning 'unsatisfiable'), so no such counterexample exists. It also exports a verifiable certificate of unsatisfiability which can be checked in polynomial time.

The proof is somewhat unilluminating, because it doesn't give any indication as to *why* your conjecture is true, just that it is.

in$\mathcal{H}$ or all areout, and (2) the minimum element of $B$ is one larger than the minimum of $A$, and is at most $7$.' $\endgroup$ – Peter Heinig Feb 18 '18 at 10:58