Let $\displaystyle A=\{a\in\mathbb R:a>0,\,a^2<2\}.$ $\displaystyle A\ne\O$ as e.g. $\displaystyle 1\in A.$ Also $\displaystyle A$ is bounded above. $\displaystyle \therefore\ x=\sup A$ exists. Note that $\displaystyle 1\leqslant x$ since $\displaystyle 1\in A.$
Suppose $\displaystyle x^2>2.$ Then $\displaystyle \frac{x^2-2}{2x}>0$ and so $\displaystyle \exists\,n\in\mathbb N$ such that $\displaystyle 0<\frac1n<\frac{x^2-2}{2x}.$ Then
$\displaystyle \left(x-\frac1n\right)^2\ =\ x^2-\frac{2x}n+\frac1{n^2}\ >\ x^2-\frac{2x}n\ >\ x^2-(x^2-2)\ =\ 2$
so $\displaystyle a\in A\ \Rightarrow\ a^2<2<\left(x-\frac1n\right)^2\ \Rightarrow\ a<x-\frac1n$ contradicting the leastness of $\displaystyle x.$
Now suppose $\displaystyle x^2<2.$ Choose a natural number $\displaystyle n$ such that $\displaystyle 0<\frac1n\leqslant\frac{2-x^2}{4x}$ and $\displaystyle \frac1n<2x.$ Then
$\displaystyle \left(x+\frac1n\right)^2\ =\ x^2+\frac{2x}n+\frac1{n^2}\ <\ x^2+\frac{2x}n+\frac{2x}n\ \leqslant\ x^2+2-x^2\ =\ 2$
so $\displaystyle x+\frac1n\in A$ contradicting the fact that $\displaystyle x$ is an upper bound for $\displaystyle A.$
Since both $\displaystyle x^2<2$ and $\displaystyle x^2>2$ lead to a contradiction, we conclude that $\displaystyle x^2=2.$