For positive $k,c,r$ there is an $N$ such that every $c$-colouring of $[N{]}^k$ has a monochromatic $r$-element set

If $r=1$ or $c=1$, any sufficiently large finite set works. For $k=1$, $N=c(r-1)+1$ works by finite pigeonhole. For $k=2$, repeatedly group one colour against all remaining colours and apply [L1]; induction on $c$ gives a finite multicolour graph witness for every target $r$.

Assume $k\ge 3$ and that the theorem is known for $(k-1)$-subsets with every finite colour and target parameter. Put $M=c(r-1)+1$. Choose finite reservoir sizes backwards by $q_M=1$ and, for $i<M$, let $q_i$ be one more than a $(k-1)$-uniform Ramsey witness for target $q_{i+1}$ and $c$ colours, which exists by the induction hypothesis.

Starting with a $q_0$-element set, choose its least vertex $x_0$. Colour each $(k-1)$-subset $A$ of the remaining reservoir by the colour of $A\cup \{x_0\}$, and restrict to a homogeneous $q_1$-element reservoir. Repeat. After $M$ stages there are vertices $x_0,\dots ,x_{M-1}$ and colours $d_0,\dots ,d_{M-1}$ such that every $k$-set of chosen vertices whose least member is $x_i$ has colour $d_i$.

Finite pigeonhole gives indices $i_1<\cdots <i_r$ for which $d_{i_1}=\cdots =d_{i_r}$. Every $k$-subset of $\{x_{i_1},\dots ,x_{i_r}\}$ has least element $x_{i_j}$ for some $j$, hence has this common colour by step 2.1. This is a monochromatic $r$-set.

The bases and the step from $(k-1)$ to $k$ prove the assertion for every positive $k,c,r$.