Worked solution: Kahn–Kalai's disproof of Borsuk's conjecture via the Frankl–Wilson theorem

A piece of $\mathcal K$ has strictly smaller diameter exactly when it contains no two cuts $S(A,B), S(C,D)$ with $|A \cap C| = k$ (the minimal-intersection, maximal-distance pairs from Step 4). Applying the Frankl–Wilson theorem of Step 3, with the ground set $V$ of size $n := m$ and forbidden intersection size $n/4 = k$, any such 'safe' sub-family of $\mathcal K$ has size at most $2\binom{m-1}{k-1}$. Since $\mathcal K$ itself has $\tfrac12\binom{m}{2k}$ elements, partitioning $\mathcal K$ into pieces of smaller diameter requires at least $\tfrac12\binom{m}{2k} \big/ 2\binom{m-1}{k-1}$ pieces.

A Stirling-approximation computation (carried out for $m=4k$ with $k$ ranging over prime powers, using the prime number theorem to guarantee prime powers are not too sparse) shows this ratio exceeds $(1.203)^{\sqrt d}$ once $d = \binom{m}{2}-1$ is large enough, which is exponentially larger than the conjectured $d+1$. Since $\mathcal K \subset \{0,1\}^d$ is a genuine bounded subset of $\mathbb{R}^d$ of diameter equal to the maximal distance computed in Step 4, this directly disproves Borsuk's conjecture for all sufficiently large $d$.