See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=283904&hilit=distance+twice. fedja's comment that "the condition that the numbers are distinct is missing" is somehow made moot by the fact that both posts speak of a set $S$ of numbers, and as such implicitely assuming they are distinct; but nevertheless it is worth making this observation - indeed, otherwise we can just take all $x_2=x_3=\cdots = x_n$ (for $n\geq 4$) to be some irrational number.

We consider the basis of $x_i$. Let that be $\mathbb{Q}[r_1,r_2,\ldots,r_m]$. Then choose any $r$ let us say $r_1$. Since $r_1$ is in the basis thus there must exist two $x_i$ such that there difference contains $r_1$ (This ensures that the component of $r_1$ in the difference we'll define isn't $0$) . Let $X_1$ = Those $x_i$ who have the maximum component of $r_1$ and $X_2$ = Those $x_i$ having minimum component of $r_1$. Then consider the difference of "rightmost" (i.e. greatest) member of $X_1$ and "leftmost" ( i.e. smallest ) member of $X_2$. If this has to have a friend then one vertex must be in $X_1$ and other in $X_2$. But if you choose any such pair (other than this one) then its length will be strictly less than the earlier chosen pair, the most preposterous statement! So all lengths must be rational (as we must dustbin any $r_i's$ if they pop up just like that quantum fluctuations anyone?) $\square$. IMPORTANT:This stunt has been performed under the supervision of biomathematics. Do not try at home. Patreon supporters: 1. (to be disclosed) 2. (to be disclosed)

Very instructive approach using Hamel basis. It so happened that I recently commented on two other problems using the same method. One may see all three of them (including this one) in my blog.