= Solution
Fix a number $k$ of colors. For a two-point set whose points are distance $d$ apart, take the vertices of a <regular simplex> with $k+1$ vertices and side length $d$. The <pigeonhole principle> gives two vertices of one color, and they form the required congruent copy. Thus every two-point set, equivalently every <line segment>, is a <Euclidean Ramsey set>.
For an <equilateral triangle> of side length $d$, take a <regular simplex> with $2k+1$ vertices and side length $d$. The <pigeonhole principle> gives three vertices of one color, and every three vertices of a regular simplex form an equilateral triangle. Hence every equilateral triangle is Euclidean Ramsey.
To prove the <product theorem for Euclidean Ramsey sets>, let $S_X$ be a finite Ramsey witness for $X$ under $k$ colors. There are at most $k^{|S_X|}$ possible color patterns on $S_X$. Choose a finite Ramsey witness $S_Y$ for $Y$ under that many colors. Given a $k$-coloring of $S_X\times S_Y$, color each $y\in S_Y$ by the complete pattern
$$
x\longmapsto c(x,y),\qquad x\in S_X.
$$
There is a copy $Y'\cong Y$ on which this pattern is constant. The common pattern on $S_X$ contains a monochromatic copy $X'\cong X$. Every point of $X'\times Y'$ then has the same original color, and the orthogonal product is congruent to $X\times Y$.
A rectangle is the <Cartesian product> of two line segments, so it is Euclidean Ramsey. Three suitable vertices of a rectangle form a <right triangle>; any subset of a monochromatic set is monochromatic. Thus every right triangle is Euclidean Ramsey.
It remains to show that the collinear set $\{0,1,2\}$ behaves differently. In every $\mathbb R^m$, use the <finite coloring>
$$
c(x)=\lfloor2\|x\|^2\rfloor\pmod {10}.
$$
A congruent copy has the form $a-v,a,a+v$ with $\|v\|=1$ for the <Euclidean norm>. The <parallelogram law> gives
$$
2\|a+v\|^2+2\|a-v\|^2-4\|a\|^2=4.
$$
Put $n_+=\lfloor2\|a+v\|^2\rfloor$, $n_-=\lfloor2\|a-v\|^2\rfloor$, and $n_0=\lfloor2\|a\|^2\rfloor$. The errors introduced by the three <floor functions> show that
$$
2<n_++n_--2n_0<6.
$$
If all three points had one color, the integer in the middle would be divisible by ten, which is impossible. This proves the <three-term unit arithmetic progression is not Euclidean Ramsey> assertion.
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