How to prove $X = \{x\in \mathbb{R} \ : \ x^2 < a \} $ has supremum? Assume that $a>0$, Suppose we have :
$$X = \{x\in \mathbb{R} \ : \ x^2 < a \}$$
We should prove that this set has a supremum, and that's $\sqrt{a}$ .
I saw this answer on one of the related posts:

Suppose that $a>0$ then $\sqrt{a}$ is an upper bound . To see this, use the definition of an open ball . Also $0 \in (-\sqrt{a},\sqrt{a})$ since $|0|<\sqrt{a}$. Therefore supremum exists. Now assume for contradiction that $\sqrt{a}$ is not the least upper bound. Then there exist $M \in R$ which is the supremum and $M<\sqrt{a}$.Consider $z:=\frac{\sqrt{a}-M}{\sqrt{a}}+M$.By construction $z>M$. it is impossible that $z<\sqrt{a}$ since M is the supremum,But if $\sqrt{a}\leq z$, then $\sqrt{a}\leq\frac{\sqrt{a}-M}{\sqrt{a}}+M \to \sqrt{a}\leq M$ ,contradiction.

My first question:
Is how author recognized that she should use $\frac{\sqrt{a}-M}{\sqrt{a}}+M$ ? Can we determine a logical process to achieve this expression for our needs?
My second question:
I have problem with this part:
$$\sqrt{a}\leq\frac{\sqrt{a}-M}{\sqrt{a}}+M \to \sqrt{a}\leq M$$
Can we conclude from $z>M$ and $\sqrt{a}\leq z$ that $\sqrt{a}\leq M$ ? I think that's not possible!
Last one:
Is there any better way to prove that?
 A: The argument I posted below some time ago is suspect. More specifically, what does $\sqrt{a}$ even mean a priori? The existence of $\sqrt{a}$ is justified by the completeness of $\mathbb R$, as the supremum of $\{x \mid x^2 < a\}$. What is there left to prove?

$\sqrt{a}$ is an upper bound of $X$, hence $\sup X$ exists. Suppose for a contradiction $\sup X < \sqrt{a}$. Then by definition of supremum, there exists $x\in X$ such that $\sqrt{a}< x$, which leads to $a<x^2$, a contradiction.
Thus, $\sup X=\sqrt{a}$ must hold.
A: 
$$\sqrt{a}\le\frac{\sqrt{a}-M}{\sqrt{a}}+M\iff\frac{\sqrt{a}-M}{1}\le\frac{\sqrt{a}-M}{\sqrt{a}}$$

This is perfectly plausible if $0<\sqrt{a}<1$. As a direct counterexample, let $a=1/4$ and $M=1/8$. $\sqrt{a}>M$ yet $\frac{\sqrt{a}-M}{\sqrt{a}}+M=\frac{7}{8}>\sqrt{a}$, so we don't get the implication "$\sqrt{a}\le M$, a contradiction." The answer to your first question is: It doesn't matter, because what you've written is wrong! I think this answers your second question too.
As for the last one: I'd prove it like this. I will also completely avoid assuming $\sqrt{a}$ even exists, I will assume only that $a\ge0$ (for, if $a<0$ then $X$ is an empty set with no (or no "meaningful") supremum).
An important point that I'll use without further comment: note that, if $x\in X$ and $0\le y<x$, then $y\in X$ as well (check this yourself).
Let $x_0=\max\{1,a\}.$ Then $x_0^2\ge\max\{1,a^2\}\ge a$ in all cases where $a\le 1$ or $a>1$. Therefore $X$ has an upper bound, $x_0$. $X$ is also nonempty since $0\in X$. Therefore there is indeed a supremum of $X$, call it $\alpha\in\Bbb R$. Necessarily $\alpha\ge0$. We want to show $\alpha^2=a$: that is equivalent to showing $\sqrt{a}$ exists and equals the supremum $\alpha$.
Then we get that, for any $\alpha>\varepsilon>0$, that $(\alpha-\varepsilon)\in X$ hence $\alpha^2-2\varepsilon\alpha+\varepsilon^2\le a$, or: $\alpha^2\le a+2\varepsilon\alpha-\varepsilon^2$. $\alpha$ is a finite number... this inequality holds for all arbitrarily small $\varepsilon$, so it must be that $\alpha^2\le a$. Likewise, for any $\varepsilon>0$, $\alpha+\varepsilon\notin X$ (otherwise $\alpha$ would not be the supremum!) so $\alpha^2+\varepsilon^2+2\varepsilon\alpha>a$. Again, this shall hold for arbitrarily small $\varepsilon$, so it must be that $\alpha^2\ge a$.
It follows that $\alpha^2=a$, so $\alpha=\sqrt{a}$ is the supremum of $X$.
You can also show that $-\sqrt{a}$ is $\inf X$. Perhaps this would be a good exercise.
A: This supposed proof has some flaws. For example, if $a > 1$, then $\frac{\sqrt{a} - M}{\sqrt{a}} < \sqrt{a} - M$, which implies that $\frac{\sqrt{a} - M}{\sqrt{a}} + M < \sqrt{a} - M + M = \sqrt{a}$, contrary to the claim made in the cited argument. So, I'd recommend looking more closely at the accepted answer from the cited post for pointers on this problem.
