I am familiar with Rudin's proof that $\sup \{ p\in\mathbb{ Q}^+\mid p^2<2 \} =\sqrt 2$ ?
Let $A$ be the set of all positive rationals $p$ such that $p^2 < 2$ and let $В$ consist of all positive rationals $p$ such that $p^2 > 2$. We shall show that $A$ contains no largest number and $В$ contains no smallest. More explicitly, for every $p \in A$ we can find a rational $q \in A$ such that $p < q$, and for every $p \in В$ we can find a rational $q \in В$ such that $q < p$. To do this, we associate with each rational $p > 0$ the number $$ \begin{align} q &= p - \frac{(p^2 - 2)}{p + 2} &(2)\\ &=\frac{2p + 2}{p + 2} &(3)\\ &\Rightarrow q^2 - 2 = \frac{2(p^2 - 2)}{(p + 2)^2} &(4) \end{align} $$ If $p \in A$ then $p^2 — 2 < 0$, (3) shows that $q > p$, and (4) shows that $q^2 < 2$. Thus $q \in A$. If $p \in В$ then $p^2 — 2 > 0$, (3) shows that $0 < q < p$, and (4) shows that $q^2 > 2$. Thus $q \in B$."
Are there alternative proofs that $\sup \{ p\in\mathbb{ Q}^+\mid p^2<2 \} =\sqrt 2$ ?
You can prove that $\mathbb{Q^+}$ is dense in $\mathbb{R^+_0}$ and then you have: $p \leq \sqrt{2},\forall p \in A$ and there is a sequence $(x_n),\ x_n \in \mathbb{Q^+}$, with property $x_n \in (\sqrt{2}-\frac{1}{n},\sqrt{2})$ (since it is dense). It is obvious that $(\forall \epsilon >0)(\exists x_n \in A)(n>\frac{1}{\epsilon})(\sqrt{2}-\epsilon< \sqrt{2}-\frac{1}{n}\leq x_n\leq\sqrt{2})$, meaning both conditions of a supremum are satisfied.