Let $A$ and $B$ be nonempty sets and let $A\cup B$ be algebraically independent set over $\mathbb Q$ and $A\cap B=\emptyset$ . Let $C\subset\mathbb Q.$ Now, fix an $r\in R$ we have, $$ r\cdot A\cdot C\not\subset B$$ where $A\cdot C=\{a\cdot c\colon a\in A\ \&\ c\in C\}.$
It is enough to show that $r\cdot A\not\subset B.$
If $r\in\mathbb Q$ then it is an impossible because $A\cup B$ is algebraically independent. Otherwise, $r\in\mathbb R\setminus\mathbb Q$, for contradiction, assume $r\cdot A \subset B.$. Then for every $x\in A$ there exists $y_x\in B$ such that $r=\frac{y_x}{x}$. Also, $r=\frac{y'_{x'}}{x'}$ where $x'\in A$ and $y'_{x'}\in B$. But this means, $r\in\mathbb Q(x,y_x)\cap\mathbb Q(x',y'_{x'})=\mathbb Q.$
Here I used the well known result: Let $E$ be an extension field of the field $F$ and suppose that $a,b\in E$ are algebraically independent over
$F$. Show that $F(a)\cap F(b) = F.$
which is a contradiction since $r\not\in\mathbb Q$. In case $y_{x}=y'_{x'}$, it also gives a contradiction.
Now, I proved $r\cdot A\not\subset B$, which means, there exists at least $x^*\in A$ such that $r\cdot x^*\neq y$ for all $y\in B.$ To complete the proof, for contradiction, assume $$ r\cdot A\cdot C\subset B$$
This means, there exist $q,q'\in\mathbb Q$ and $y_q,y'_{q'}\in B$ such that $r\cdot x^*\cdot q=y_q$ and $r\cdot x^*\cdot q'=y_{q'}$. But this impossible because $$r\cdot x^*(q'y_q-q y_{q'})=0$$ since $rx^*\neq 0$ so $q'y_q-q y_{q'}=0$ which is contradiction since $\{y_q, y_{q'}\}$ is algebraically independent. This finihes the proof.
My question, Does the prove above correct? Is there is short way to prove that?
Any help will be appreciated greatly.