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Put $h=\sqrt3/2$ and color $(u,v)$ by $\lfloor v/h\rfloor$ modulo two. A unit equilateral triangle has vertical span at least $h$: one of its three unoriented edge directions is within $30$ degrees of vertical. Its span is at most $1$. Hence a monochromatic triangle cannot lie in one half-open strip and can meet at most two strips of its color. In the latter case the lone vertex in one strip has vertical displacement strictly greater than $h$ from each of the other two vertices, in the same direction. The two unit vectors to those vertices make angle $60$ degrees, so their sum has length $\sqrt3$. Its vertical component would exceed $2h=\sqrt3$, a contradiction. The coloring avoids all unit equilateral triangles, including boundary cases.

Suppose the statement of L32 fails for some $B$. There are then positive $x_j\to1$ and planar sets of at most $B$ points whose hypergraphs of equilateral triples of lengths $1$ or $x_j$ are not two-colorable. Some component of each hypergraph is not two-colorable, since separate component colorings combine. Pass to these components and a subsequence with fixed number of labeled vertices and fixed abstract hyperedge set; there are only finitely many possibilities.

Translate one vertex to the origin. In a connected component, any vertex can be reached by a path of at most $B-1$ pairs belonging to hyperedges. Those pairs have length at most $2$ eventually. All coordinates are therefore bounded, and a subsequence converges coordinatewise. Every hyperedge limits to three distinct points forming a unit equilateral triangle. Vertices outside a common hyperedge may coalesce; this does not matter. Pull back the strip coloring of the limiting points to the labels. It properly colors every hyperedge of the fixed abstract hypergraph, contradicting its non-two-colorability.