 Solution

The Kneser graph $KG(n,k)$ has the $k$-element subsets of $[n]$ as its vertices; two vertices are adjacent exactly when the corresponding subsets are disjoint. The Lovász theorem on Kneser graphs determines its chromatic number. A graph colouring is therefore a partition of the $k$-sets into intersecting families.

First construct a graph colouring with $d+2=n-2k+2$ colours. If a $k$-set meets $[d+1]$, give it the colour of its least element. Give every remaining $k$-set colour $d+2$. Two sets with one of the first $d+1$ colours share that colour's element. The last colour consists of $k$-sets in a $(2k-1)$-element ground set, so it too is an intersecting family. Thus $\chi (KG(n,k))\le d+2$.

For the converse we establish the Gale hemisphere lemma explicitly, using a signed moment curve. Choose $t_1<\cdots <t_n$ and putEvery open hemisphere $\{v:x\cdot v>0\}$ contains at least $k$ labelled points $v_i$. To see this, put $p(t)=x_0+x_{1}t+\cdots +x_d{t}^d$. It is a nonzero polynomial of polynomial degree at most $d$. First suppose none of the $p(t_i)$ vanishes. Write $b_i=\mathrm{sgn}((-1{)}^{i}p(t_i))$. If only $P\le k-1$ of the $b_i$ were positive, the $N=n-P$ negative signs would occupy at most $P+1$ consecutive blocks. There would be at leastadjacent pairs with both $b_i,b_{i+1}$ negative. Each such pair forces $p(t_i)$ and $p(t_{i+1})$ to have opposite signs. The intermediate value theorem would give $d+1$ distinct roots of a polynomial, a contradiction.

If $p$ vanishes at $z$ of the sample points, then $z\le d$. A Lagrange interpolation polynomial $q$ of polynomial degree at most $z-1$ can be chosen with $(-1{)}^{i}q(t_i)<0$ at all those points. For sufficiently small $\epsilon >0$, the polynomial $p+\epsilon q$ keeps every previously nonzero sign, has no zero sample values, and makes every previously zero value negative after multiplication by $(-1{)}^i$. Its positive count is exactly the positive count of $p$, so the preceding argument proves the open hemisphere assertion also in this case. For $d=0$, the labelled points alternate between the two points of $S^0$; distinct labels, rather than distinct positions, are what is needed.

Suppose now that $KG(n,k)$ had a graph colouring with at most $d+1$ colours, padding the palette with unused colours if necessary. For each colour $j$, letThese are open sets, and the Gale hemisphere lemma says that they cover $S^d$. If both $x$ and $-x$ belonged to $U_j$, two $k$-sets of colour $j$ would lie in opposite open hemispheres. They would be disjoint, hence adjacent in the Kneser graph, contradicting the graph colouring. The Lusternik-Schnirelmann-Borsuk theorem excludes this cover. Therefore the lower bound matches the explicit colouring: