Consider the set $T$={$t\in\Bbb{R}:t^2<2$}
Let $\alpha=supT$ which implies $\forall\epsilon>0\exists t\in T$ such that $\alpha-\epsilon<t$.
Suppose $\alpha^2<2$ $\Rightarrow$ $\alpha^2=2-\epsilon_0$, for some $\epsilon_0\in\Bbb{R}$. By the Archimedian Property, for every $\epsilon_0$ there exist $n\in\Bbb{N}$ such that $\frac{1}{n}<\epsilon_0$. Let us let $\epsilon_1=\frac{1}{n_1}<\epsilon_0$ for some $n_1\in\Bbb{N}$. Then there exist $t_1\in T$ such that $t_1^2=2-\epsilon_1$ which implies that $t_1^2=2-\epsilon_1>\alpha^2=2-\epsilon_0$ . By property of a supremum, this cannot be because $t_1\in T$. Hence, a contradiction. Therefore, it cannot be that $\alpha^2<2$
Suppose $\alpha^2>2$ $\Rightarrow$ $\alpha^2=2+\epsilon_2$ where $\epsilon_2>0$. Again by the Archimedean property, we can find a real number $\epsilon_3=\frac{1}{n_3}$ for some $n_3\in\Bbb{N}$ such that $\epsilon_3<\epsilon_2$, $\forall\epsilon_2\in\Bbb{R}$. Let $q=\epsilon_2 - \epsilon_3$. Then $\alpha^2-q=2 + \epsilon_3$ which means $(supT)^2-q>t, \forall t\in T$. Thus contradicting what a supremum is.
Therefore, $\alpha^2$ cannot be greater or less than $2$. Meaning $\alpha^2=2$
Consider taking the set $S = \{r \in \mathbb{R} | r^2 > 2\}$
Then show that $\alpha = \sup(T)$ is not in $S$ or $T$ which means $x^2$ is neither greater than or less than than $2$.
To finish the proof, first, assume $\alpha \in S$. Then $\alpha ^2 > 2 \implies (\alpha-\epsilon)^2 >2$ for sufficiently small $\epsilon \implies t \leq \alpha-\epsilon < \alpha$ for $t \in T$ which contradicts $\sup$.
Assume $\alpha \in T$. Then $\alpha^2 < 2 \implies (\alpha + \epsilon)^2 < 2$ for sufficiently small $\epsilon \implies \alpha + \epsilon \in T$ but $\alpha < \alpha + \epsilon$ which is a contradiction.