Consider two sets $A,B$ composed of two real numbers each.
These four real numbers are in $[0,1]$.
Consider other two real numbers $c\in [0,1]$, $d\in [0,1]$.
Assume there exists a way of ordering the two numbers in each set $A,B$ such that $$ \begin{cases} w^A_1+w^B_1=c\\ w^A_2+w^B_2=d \end{cases} $$ where
$w^A_h$ denotes the $h$th element of set $A$ once we have ordered its two elements
$w^B_h$ denotes the $h$th element of set $B$ once we have ordered its two element
Claim: if such an ordering is not unique, then it should be that the two numbers in $A$ are equal and/or that the two numbers in $B$ are equal.
Is this claim correct? If yes, how can I prove it? If not, can you provide a counterexample?
Say the ordering is not unique. Then at-least one of $A,B$ can be ordered in two ways. Without loss of generality, let us assume that set is $A$. Keeping the order of $B$ intact, we have
$\implies w_1^A+w_1^B=w_2^A+w_1^B=c\implies w_1^A=w_2^A$
Therefore, both the elements of $A$ are identical. We can assume that the ordering of $B$ is not unique and land at a similar conclusion for $B$.
Edit. As suggested by the OP, I considered the case when the ordering of either $A$ or $B$ was not unique, but forgot to consider the case when both their orderings changed. In that case, $c=d$ and the elements of $A,B$ need not be equal.