Show that $\mathbb{Q}$ is dense in the real numbers. (Using Supremum)

Another way to see that that $\mathbb Q$ is dense in $\mathbb R$ is by showing that $\frac{\left\lfloor{nx}\right\rfloor}{n}$ is a sequence of rational numbers that converges to $x$ for $n$ goes to infinity. If you want this sequence also helps you to show that $x$ is the supremum of $\mathbb Q_x$, because $\frac{\left\lfloor{nx}\right\rfloor}{n}$ converges to $x$ from below...

To be clear the floor function is defined as $\left\lfloor{y}\right\rfloor:= \max \{m\in \mathbb Z: m \leq y \}$.

By definition of supremum, note that for any $\epsilon>0$ we have that $x-\epsilon$ is not an upper bound for $\Bbb Q_x$, meaning there is some $q_\epsilon\in\Bbb Q_x$ such that $x-\epsilon<q_\epsilon$, so $x-q_\epsilon<\epsilon$ as desired.

If you want to prove it for $x<0$, try applying a similar argument to $\Bbb Q_{-x}$. That should allow you to find a $q_\epsilon$ that is greater than $x$, and within $\epsilon$ of it.