Simple supremum/infimum proof Let $A$ and $B$ be non-empty  subsets of $\mathbb{R}$ with  $a\leq b$ for all $a\in  A, \space b \in B$ and suppose there exists a  unique number $\alpha$ such that $a\leq \alpha \leq b$ for all $a\in  A, \space b \in B.$ Then we have $\alpha  = \sup A = \inf B$.
Proof: Suppose $\alpha < \sup A.$ If $S = \sup A\in  A$ we are done so  assume  this is not the case. Choose  $k = \dfrac{\alpha +  S}{2}$ so $\alpha < k$ and $k < S.$ Now we can choose some $A\ni  a \geq k$ which yields a contradiction  since $\alpha <a$. One can similarly show that $\alpha \leq \inf B.$ We conclude that  $\sup A \leq \alpha \leq \inf B$ and since $\alpha$ is unique it follows that $\alpha  = \sup A = \inf B$.
Does my proof seem sound? If  not, what do I have to adjust?
 A: Your argumentation is correct but I would add a bit more detail to the last sentence: Assume that $\sup A\neq \alpha \neq \inf B$ so we would obtain the inequality $\sup A < \alpha < \inf B$. Then we could choose a number
$t = \dfrac{
\alpha + \inf B}{2}$ (analogous to how you chose $k$ in your proof) so $\sup A < \alpha  < t < \inf B$ which would contradict the uniqueness of $\alpha.$ The same applies for $\sup A$. This makes your last conlcusion more clear in my opinion.
In general, it's good to try keeping the proof concise whereby one should not leave out details that make it "hard" to follow.
A: I was initially lost by your first two sentences. It is not super clear with what "we are done" if $\sup A$ belongs to $A$. In any case, we can take the given existence of $\alpha$
and apply just the definitions of supremum and infimum to say $\sup A \leq \alpha \leq \inf B$.
Your final statement is sound, although it took me some time to become convinced. Of course it is intuitively obvious. You might want to add some detail to make clear why "it follows".   As for myself,
I became convinced when I wrote it this way: $\,\,\,\,\alpha$ is unique $\,\,\implies\,\, \inf B \,\leq\, \sup A.$
Or, even more clear to me is: $\,\,\,\,\alpha$ is not unique  $\,\,\impliedby\,\, \sup A \,< \,\inf B.$
