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Source: original paper, printed p. 539, Theorem 5.

Statement

Let $S\subset\mathbb R^m$ have $N$ points. Suppose every subset of $q=\lfloor N/r\rfloor$ points contains a congruent copy of a given nonempty configuration $K$, where $r>0$ and $1\le q\le N$. Let $L\subset\mathbb R^m$ have $l$ points with $l<r$. Every red-blue coloring of $\mathbb R^m$ has a red congruent copy of $K$ or a blue translate of $L$.

Full proof

Assume there is no red copy of $K$. For every $a\in L$, the translated witness $a+S$ has fewer than $q$ red points. Thus fewer than $q$ choices of $s\in S$ make $a+s$ red. The union over all $l$ choices of $a$ excludes at most $l(q-1)$ choices of $s$. In particular,

$$l(q-1)\le lq\le lN/r<N.$$

Some $s\in S$ is excluded by none of them. Every point of $L+s$ is blue, as required. If $L$ is empty, its translate is already blue and the same conclusion is immediate.

Only a finite family of bad-choice sets is counted. Distinct translates may overlap without affecting the argument. The copy of $L$ is a translate, while the red conclusion only requires congruence.

Used by. Corollary 6.