Let $a,b,c$ are lengths of sides of the triangle $ABC$. Suppose the contrary. Then the following lemma should be true: $Lemma:$ If $PQ$ is a segment and $|PQ|=a$ or $b$ or $c$, then points $P$ and $Q$ cannot be both red. $Proof$ $of$ $Lemma:$ Suppose the contrary, $P$ and $Q$ are red. Select a point $R$ from plane which satisfies for a permutation $(X,Y,Z)$ of $\{A,B,C\}$, $PQR$ is congruent to $XYZ$. Consider circles $c_1,c_2,c_3$ with radius $1$ and whose centers are $P,Q,R$, respectively. Then each point from $c_1$ and each point from $c_2$ is green. If all points on $c_3$ are red, then there will be two red points on $c_3$ with the distance $1$, so there is a green point $M$ on $c_3$. Then there should be two points $K,L$ on $c_1,c_2$ respectively which satisfy $KLM$ is congruent to $XYZ$. Since $K,L,M$ are all green, this is a contradiction. $\Box$ Let $a\geq b\geq c$. Select a red point $P$ from plane. Consider the circle $C$ which its center is $P$ and length of its radius is $a$. Then all points on $C$ should be green. Select two points $K,L$ on $C$ with the distance $c$. Then we can select a point $M$ from outside of the circle $C$, which satisfies $|KM|=a$ and $|LM|=b$. Then $M$ should be red. Now consider points $K',L',M'$ which satisfy $P$ is the midpoint of the segments $KK'$, $LL'$ and $MM'$. Then $K',L'$ are green, $M'$ is red and $K'L'M'$ is congruent to $KLM$. We can get the triangle $K'L'M'$ by rotating $KLM$ on circle $C$ $180$ degrees. Since $|M'M|>2a$, there should be a triangle $K''L''M''$ which we can get by rotating $KLM$ on the circle $C$, and which satisfies $|M''M|=a$. Since this contradicts with the lemma, the proof is done.

nice proof, but I think you could make it shorter; it seems me that there's no benefit from the construction of K',M',L'.