Show that every irrational number in <mrow class="MJX-TeXAtom-ORD">

Let $\alpha$ be irrational. For each positive integer $n$ there is a least integer $k$ such that ${\displaystyle \frac{k}{n}}>\alpha$, and then this number is rational. You need to prove that this sequence tends to $\alpha$.
Now let $\alpha$ be rational, and repeat the above argument with something like ${\displaystyle \frac{k\sqrt{2}}{n}}$ instead of ${\displaystyle \frac{k}{n}}$.
[You'll also need to prove that ${\displaystyle \frac{k\sqrt{2}}{n}}$ is irrational whenever $k\ne 0$, and cook up a way of avoiding having $0$ in the sequence.] Edit: Or consider $\alpha +\frac{\sqrt{2}}{n}$ or something like that, as Cameron Buie suggests in the comments.
The moral of the story is that $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$.