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

Let $V = \{1, \ldots, m\}$ with $m = 4k$ for $k$ a prime power, and let $W$ be the set of all $\binom{m}{2}$ unordered pairs of elements of $V$ — the edges of the complete graph on $V$ — so $d := |W| = \binom{m}{2}$ is the ambient dimension. For every partition $\{A,B\}$ of $V$ into two halves of size $2k$ each, let $S(A,B) \subset W$ be the set of pairs with one element in $A$ and one in $B$ (the 'cut' between $A$ and $B$); note $|S(A,B)| = (2k)^2 = m^2/4$ for every such partition. Let $\mathcal K = \{S(A,B) : \{A,B\}$ a balanced partition of $V\}$, a family of $\binom{m}{2k}/2$ subsets of $W$, each identified with a $(0,1)$-vector in $\mathbb{R}^d$.

Because every set in $\mathcal K$ has exactly the same size $m^2/4$, all points of $\mathcal K$ lie on a common sphere in $\mathbb{R}^d$ centred at the all-$\tfrac12$ point, and by Step 2's distance formula their pairwise distances are governed entirely by $|S(A,B) \cap S(C,D)|$: two cuts are close together exactly when their crossing-edge sets overlap a lot. The smallest overlap — and hence the largest distance, the one realising $\mathcal K$'s diameter — turns out (by direct computation with the inclusion–exclusion of $A, B, C, D$) to occur exactly when $|A \cap C| = k$, setting up the intersection condition that Step 3's theorem is built to control.