Question on a proof that there exist $\alpha \in \mathbb R$ such that $\alpha^2=b$ for $b \in \mathbb R$ and $b\geqslant0$ I am aware there are a couple of other questions related to this topic. I could formulate a proof but I have a particular question regarding a lecturer's proof I found in the web.
The proof goes like this:
Define $T=\{t\in\mathbb R\mid x^2<b\}$. Let $\alpha=sup \: T$. We have to prove that $\alpha^2=b$ by ruling out $\alpha^2<b$ and $\alpha^2>b$. I will omit the case $\alpha^2<b$ since it was clear for me.
For the case $\alpha^2>b$, let's write:
$\left( \alpha - \frac{1}{n} \right)^2 = \alpha^2 - \frac{2\alpha}{n} + \frac{1}{n^2} > \alpha^2 - \frac{2\alpha}{n}$ where $n \in \mathbb N$
Then, we pick a $n_0 \in \mathbb N$ satisfying $\frac{1}{n_0}<\frac{\alpha^2-b}{2\alpha}$ or $\frac{2\alpha}{n_0}<\alpha^2-b$. We know $n_0$ exists from the archimedean property. Then we would have:
$\left( \alpha - \frac{1}{n_0} \right)^2 > \alpha^2 - \frac{2\alpha}{n_0}>\alpha^2-(\alpha^2-b)=b > t^2 \: \: \: \forall t \in T  \tag{1}\label{1}$
From the last inequality, the proof concludes that $\left( \alpha - \frac{1}{n_0} \right)$ is an upper bound of $T$ which is less that $\alpha$, then contradicting $\alpha=sup \: T$.
And here is my reflection and question. I think that the only thing we can conclude from the inequality $\eqref{1}$ is that $\left( \alpha - \frac{1}{n_0} \right) \notin T$.
If $\left( \alpha - \frac{1}{n_0} \right)>0$, then the conclusion would hold true since $\left( \alpha - \frac{1}{n_0} \right) > t \: \: \: \forall t \in T$.
However, it can also be true that $\left( \alpha - \frac{1}{n_0} \right)<0$ and in this case I cannot find any contradiction.
Am I right or am I missing something? and How can I finish the proof from $\eqref{1}$?

I think the proof is a bit sloppy. You can start by considering the case $b=0$. In that case $\alpha=0$ does the job. If $b>0$ your set $T$ is not empty. You can also notice that $T$ is symmetric, that is whenever $t\in T$, $-t\in T$. Therefore $\alpha\ge 0$.
In case $\alpha^2>b$ you necessarily have $\alpha>0$. Then, when you choose $n$ you can also require $1/n<\alpha$ (Archimedean property) and then the proof works.
