The significance of the set $S:=\{x\in{\mathbb Q}_{>0}\ |\ x^2<2\}$ is the following: It is a nonempty subset of ${\mathbb Q}$ which is obviously bounded above, but which has no least upper bound (called supremum) in ${\mathbb Q}$. As a consequence, the ground set ${\mathbb Q}$ is not "order complete" and should better be replaced by a more encompassing set of numbers, which of course is ${\mathbb R}$.
In order to show that $S$ has no supremum in ${\mathbb Q}$ one has to show that no number $c\in{\mathbb Q_{>0}}$ qualifies as supremum of $S$.
Given any trial $c\in{\mathbb Q_{>0}}$ then we all know that $c^2\ne2$, whence either $c^2<2$ or $c^2>2$. The "tricky" number
$\xi:={2c+2\over c+2}\in{\mathbb Q}_{>0}$
satisfies
$\xi -c={2-c^2\over c+2}\ ,\qquad \xi^2-2={2(c^2-2)\over(c+2)^2}\ .$
Now, if $\ {\rm (a)}\ c^2<2$ then it follows that $\xi>c$ and $\xi^2<2$, whence $\xi\in S$, so $c$ is not an upper bound for $S$.
If $\ {\rm (b)} \ c^2>2$ then $\xi and $\xi^2>2>x^2$ for all $x\in S$. As $t\to t^2$ is strictly increasing for $t>0$ it follows that $\xi>x$ for all $x\in S$, whence $\xi$ is an upper bound for $S$ strictly smaller than $c$.
It follows that in both cases (a) and (b) the number $c$ does not qualify as a supremum for the set $S$.
I don't think there is a systematic procedure to "invent" such a $\xi$ (after all, it is not uniquely determined). You have to fiddle around with inequalities until you hit the expression which is "just right".