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Source: original paper, printed p. 532, Theorem 2.

Statement

Every red-blue coloring of $\mathbb R^3$ has a red triangle with side lengths $1,1,\sqrt2$ or a blue unit square. The blue square can be replaced by any rectangle with side lengths $1,s$, where $0<s\le1$.

Full proof

If there is a red triangle with side lengths $1,1,\sqrt2$, we are done. Assume there is no such red triangle. If there is no red point, the blue conclusion is immediate. Otherwise choose a red point $a$. If $a$ has no red unit neighbor, its unit sphere is entirely blue. That sphere contains the desired rectangle: after translating $a$ to zero, the four points

$$(\pm1/2,\ \pm s/2,\ \sqrt{3-s^2}/2)$$

lie on it and form a rectangle with sides $1,s$.

Otherwise choose a red point $b$ with $|a-b|=1$. Let $C_a,C_b$ be the unit circles centered at $a,b$ in the planes perpendicular to $b-a$. If a point $c$ on either circle were red, the points $a,b,c$ would form a red triangle with sides $1,1,\sqrt2$. Thus both circles are blue.

Choose unit vectors $u,v$ perpendicular to $b-a$ with $|u-v|=s$. They exist in that two-dimensional perpendicular plane, since $0<s\le1<2$. Then

$$a+u,\quad a+v,\quad b+v,\quad b+u$$

are blue and form a rectangle: one side is $v-u$, the other is $b-a$, they are perpendicular, and their lengths are $s,1$. Taking $s=1$ gives the square.

If there is no red unit pair, in particular there is no red right unit triangle, so this proves the blue-square conclusion in dimension three. It does not establish the planar assertion in Problem 214.