If $u > 0$ is any real number and $x < y$, show that there exists a rational number $r$ such that $x < ru < y$.
Since $u > 0$ and $x < y$, then we know that $x/u<y/u$.
By the Density Theorem, we know there exists $r \in\mathbb Q$ such that $x/u<r<y/u$.
We finish the proof by multiplying the inequalities by $u$.
Is this enough for the proof?