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

Identify a $(0,1)$-vector $x \in \{0,1\}^n$ with the set $A \subset \{1,\ldots,n\}$ of coordinates where $x$ equals $1$. If two such vectors $x, y$ both have exactly $k$ ones (corresponding to sets $A, B$ of size $k$), then $\|x-y\|_2^2 = 2(k - |A\cap B|)$: the squared distance shrinks exactly as the overlap $|A \cap B|$ grows, and is maximised precisely when $|A \cap B|$ is as small as possible. So among $k$-subsets of $\{1,\ldots,n\}$, the pairs at maximum (diameter-realising) distance are exactly the pairs with minimum intersection.

Larman observed that Borsuk's conjecture, restricted to finite sets of constant-weight $(0,1)$-vectors, reduces to a clean combinatorial statement: if $\mathcal{K}$ is a family of $k$-subsets of $\{1,\ldots,n\}$ such that every two members share exactly $t$ elements (so $\mathcal K$ has just one non-trivial pairwise distance, realising the diameter), can $\mathcal K$ always be split into $n$ parts so that within each part every two members share at least $t+1$ elements (i.e. lie at a strictly smaller distance)? A counterexample to this purely finite question — a family $\mathcal K$ that provably needs far more than $n$ parts — is exactly what Kahn and Kalai go on to build, without even requiring the constant-intersection hypothesis in its cleanest form.