Consider an ordered field $F$ containing the rational numbers $\mathbb Q$, without assuming the completeness axiom. In particular, this means that the integers may well be bounded. (see this nice example by Tim Gowers: The unboundedness of the integers), so that we cannot use density of the rationals either, as that depends on the Archimedean principle, which is a consequence of the completeness axiom.
Consider now the set $S=\lbrace x\in {\mathbb Q}: x^2<2\rbrace$. In this context, I would like to see a proof that if $M\in F$ is any upper bound for $S$, then $M^2\geq 2$.
In fact this is false in general! Suppose for example that $F$ is a non-Archimedean extension of $\mathbb{R}$. Then $F$ has an infinitesimal element $\epsilon$, and the element $M=\sqrt{2}-\epsilon$ is an upper bound of your set $S$ (since in the definition of $S$ we only look at rationals) which has $M^2<2$.