Rudin's proof on the Analytic Incompleteness of Rationals [duplicate]

In Rudin's classical "Principles of Mathematical Analysis," he gave a proof like this:

Claim: Let $A= \{p\in \mathbb{Q} | p>0, p^2 <2\}$. Then A contains no largest number.

Proof: Given any $p\in A$. Let $q = p-\frac{p^2 -2}{p+2}$.

Later Rudin claimed that $q^2<2,$ and $q>p$. My instructor asks us to think about a question on our own: Why is such $q$ a natural choice in this proof?

I can see that in this way, $q>p$ is for sure. However, how does it become a natural choice?

I've always thought Rudin is kind of lame here. He has a fondness for pulling rabbits out of a hat, and this is not the last time you will see it.

It seems to me easier to consider

$$(p+1/n)^2 <2,\, n=1,2,\dots $$

The intuition is then clear: Surely this will be true for large enough $n,$ let's go find one, call it $n_0,$ and then $p+1/n_0$ does the job.

The idea is that as $p \to \sqrt{2}^-$, you want to add something scaling like $2-p^2$ to $p$, so that what you add goes to zero as $p \to \sqrt{2}^-$. But you can't just add $2-p^2$ to $p$. Consider for instance $p=0$, then $p+2-p^2=2$ which is too big.

How much is it too big by? Well, $p^2-2=(p+\sqrt{2})(p-\sqrt{2})$, so it is too big by a factor of $p+\sqrt{2}$. So it would be enough to divide it by any rational number greater than $p+\sqrt{2}$. $p+2$ is just what you get when you use the trivial estimate $\sqrt{2}<2$. Numerous other options would have worked, though, for example $q=p+\frac{2-p^2}{4}$.