Let $A,B\subset\Bbb R$ algebraically independent sets over $\Bbb Q$ and $A\cap B=\emptyset.$ Let $T$ be a transcdental basis of $\Bbb R$ over $\Bbb Q$ such that $A,B\subset T.$ Now, $qA$ is algebraically independent over $\Bbb Q$ for any $q\in\Bbb Q\setminus\{0\}$, easy to see.
Moreover, $qA\cap B=\emptyset.$ assume not, that is, $qA\cap B\neq\emptyset.$ So, there exists $qa=b$, where $a\in A$ and $b\in B$. So, $f(x,y)=qx-y$ is a nonzero polyoinmal such that $f(a,b)=0$ which a contradiction since $\{a,b\}\subset A\cup B.$ Is that right?