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Source. Published p. 343, Proposition 4 (published scan). Proposition 4 prints only the monochromatic form; the proof of Theorem 28 on p. 362 invokes it for the few-color form below.

Statement. Let $B$ be a set and let each member of a family $\mathcal F$ be a finite subset of $B$. If every $r$-coloring of $B$ has a monochromatic member of $\mathcal F$, some finite subfamily already has this property. Its union is a finite witness set. The same conclusion holds when a member is required to use at most a fixed number $\ell$ of colors.

Complete proof relative to compactness. Use the standard compactness of a product of finite discrete spaces, explicitly identified in external_inputs. The space of all colorings is $\Omega=\{1,\ldots,r\}^B$. For each finite $F\in\mathcal F$, let $C_F$ consist of colorings for which $F$ is not monochromatic. This condition depends on finitely many coordinates, so $C_F$ is closed. If every finite subfamily had a coloring avoiding all its members, the closed sets $C_F$ would have the finite intersection property. Compactness would give a coloring in their whole intersection, contrary to the assumption. Consequently finitely many $C_F$ already have empty intersection, as required. Their union $B'$ is finite; every coloring of $B'$ extends to $B$, so it forces a member within $B'$.

For the few-color version replace “not monochromatic” by “uses more than $\ell$ colors.” This is again a closed condition on finitely many coordinates, and the identical finite-intersection argument applies. Empty family members, if allowed, force the conclusion immediately. $\square$

The source cites a logic text for the compactness step. This page proves the precise reduction used later; it does not claim a new proof of the ambient set-theoretic compactness theorem.

Bears on. #174.