Proof step motivation in proving that set of rationals whose square is less than 2 has no upper bound

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I am flaring up my old mathematical hobby and picked up Rudin for reading. My motivation is to read with complete understanding, and thus I am trying to motivate all steps of all propositions laid out.

On page 2 it is stated (not-cited) :

Set of rationals whose square is less than 2 does not contain a greatest element

The first idea of the proof which came to my mind is to use the fact that between any two rationals, there is another rational. I proceeded by writing down $A = \{q \in \mathbb{Q} \vert q^2 < 2\}$ and $l \in A$. Now I wanted to consider the gap between number $l$ and $2$, and then any smaller gap should lead me to a rational which is less than $2$ as in:

$$l^2 < 2 \implies 0 < 2 - l^2 $$

Thus for any integer $m \geq 2$ it will hold that $l^2 + \frac{2-l^2}{m}$ is going to be a rational. The problem is that I can not construct proper $m$, so that :

$$(l+\frac{2-l^2}{m})^2 < 2$$

My problem with understanding is that in the book $m$ is simply said to be $l+2$, which does work, but I do not understand how to arrive at such conclusion that proper choice for $m$ is $l+2$

I would be grateful for your assistance.

EDIT: Rudin as in "Principles of Mathematical Analysis" by Walter Rudin