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Source. Theorem 7 and its following remark, printed p. 345, physical p. 5 of the published paper. The source credits the argument to S. Burr.

Let $C_2$ be the four vertices of a unit square. Then

$$R(C_2,6,2)\ \text{is true}. \tag{1}$$

The graph lemma

Every red-blue coloring of the edges of $K_6$ has a monochromatic $4$-cycle. To prove this, choose a vertex $v$ with at least three incident edges of one color, say red.

If $v$ has at least four red neighbors, the red edges among any four of them form a matching: two sharing a vertex would combine with $v$ to make a red $4$-cycle. The complement of a matching on four vertices contains a blue $4$-cycle, a contradiction. Thus, in the only remaining case, $v$ has exactly three red neighbors $a,b,c$ and two blue neighbors $x,y$.

The red edges among $a,b,c$ again form a matching, so after relabeling $ab,bc$ are blue. Each of $x,y$ has at most one red neighbor among $a,b,c$, since two would make a red $4$-cycle through $v$. Hence each has at least two blue neighbors there. If they have two common blue neighbors, those four vertices form a blue $4$-cycle. Otherwise their blue-neighbor sets are, after relabeling, $\{a,b\}$ and $\{b,c\}$. Then

$$v,x,b,y,v \tag{2}$$

is a blue $4$-cycle. This proves the lemma.

Euclidean realization

Let $e_1,\ldots,e_6$ be the standard basis of $\mathbb R^6$. For every edge $\{i,j\}$ of $K_6$, put

$$p_{ij}=\frac{e_i+e_j}{\sqrt2}. \tag{3}$$

A two-coloring of $\mathbb R^6$ colors these fifteen points and hence the edges of $K_6$. Let $ij,jk,k\ell,\ell i$ be a monochromatic $4$-cycle. Then

$$p_{ij},\ p_{jk},\ p_{k\ell},\ p_{\ell i} \tag{4}$$

are monochromatic. Consecutive differences in (4) have two nonzero coordinates, each of magnitude $1/\sqrt2$, so have length $1$. Consecutive side vectors use disjoint coordinate pairs and are orthogonal; opposite side vectors are negatives. Thus (4) is a unit square, proving (1).

The planar counterexample in the source

Color $(s,t)\in\mathbb R^2$ by the parity of $\lfloor t\rfloor$. Suppose a unit square with orthonormal side vectors $u,w$ were monochromatic. Along each edge the vertical change has absolute value at most $1$, so equal parities at its endpoints force the two floor values to be equal. Connectivity forces all four vertical coordinates into one half-open unit interval, whose diameter is strictly below $1$. On the other hand their vertical span is

$$|u_2|+|w_2|\ge\sqrt{u_2^2+w_2^2}=1, \tag{5}$$

because $u,w$ are orthonormal. This contradiction proves $R(C_2,2,2)$ false, including points on stripe boundaries. The paper's 1973 remark that dimensions $3,4,5$ were then undecided is historical context, not a current-status claim.