Showing $\mathbb{Q}$ does not have the L.U.B. property

If you substitute 4 for 2 in your proof, then the conclusion will be false: $\{r \in \mathbb{Q} : r > 0$ and $r^2 < 4 \}$ does have a least upper bound in $\mathbb Q$ (because 4 is a perfect square). So somewhere you have to use the fact that if $r \in \mathbb Q$, then $r^2 \ne 2$.

So your proof should have two parts:

1. Show that if $r \in \mathbb Q$, then $r^2 \ne 2$.
2. Show that if $r \in \mathbb Q$ and $r^2 > 2$, then $\exists t \in \mathbb Q$ such that $2 < t^2 < r^2$.

If you can prove these (can you?), then the result follows: any upper bound $r$ will satisfy $r^2 > 2$ (because of 1), so it can't be a least upper bound (because of 2).

Edited to reply to comment: OK, if $r^2 > 2$, then $r^2 = 2 + \epsilon$ for some rational $\epsilon > 0$. Now look at $t = r - \epsilon/2r$.

If $t^2 > 2$, then $t^2-2>0$ so there must be a positive integer $n>1$ so that $t^2-2>\frac{1}{n}$.

But $$\left(t-\frac{1}{2nt}\right)^2 = t^2-\frac{1}{n} + \frac 1{4t^2n^2} > t^2-\frac{1}{n} > 2$$

So $t-\frac{1}{2nt}$ is smaller than $t$ and it's square is still greater than 2.