The question was like this. "Prove that there exists an irrational number $\\a$ such that $x< au< y$ if $x< y$ and $u> 0$" This is my proof.
Let $a=\sqrt{2}$ which is irrational. Since $x<y$, we have $\frac{x}{\sqrt{2}}<\frac{y}{\sqrt{2}}$. By density of rational number, there exists a rational number $u>0$ such that $\frac{x}{\sqrt{2}}<u<\frac{y}{\sqrt{2}}$. Thus we have $x<\sqrt{2}u<y$ as desired.
Is my proof legit?
You're asked to find $a$ in terms of $x$, $y$, and $u$, so you can't just pick a value for $u$. Here's my version.
Choose any $x, y, u \in \mathbb R$ such that $x < y$ and $u > 0$. Then we know that $\frac{x}{u\sqrt 2}, \frac{y}{u\sqrt 2} \in \mathbb R$ and $\frac{x}{u\sqrt 2} < \frac{y}{u\sqrt 2}$. By the density of $\mathbb Q$ in $\mathbb R$, we know that there exists some $q \in \mathbb Q$ such that: $$ \frac{x}{u\sqrt 2} < q < \frac{y}{u\sqrt 2} \iff x < (q\sqrt 2)u < y $$ But then since $q \in \mathbb Q$ and $\sqrt 2 \in \mathbb R - \mathbb Q$, we may take $a = q\sqrt 2 \in \mathbb R - \mathbb Q$, as desired. $~~\blacksquare$