The following question is from Exercise 3.3.9. from Introduction to Real Analysis, Bartle and Sherbert.
Question. Let $A$ be an infinite subset of $\mathbb{R}$ that is bounded above and let $u:=\sup A$. Show there exists an increasing sequence $(x_n)$ with $x_n \in A$ for all $n \in \mathbb{N}$ such that $u=\lim (x_n)$.
Here is my proof.
Proof. Define the sequence $(x_n)$ such that $x_1 := \min \{A \}$, $x_2 := \min \{ A \setminus \{x_1\}\}$, $x_3 := \min \{ A \setminus \{x_1 , x_2\}\}$ and in general, $x_n := \min \{ A \setminus \{x_1, x_2, x_3, \dots , x_{n-1}\}\}$. $(x_n)$ has been defined in a way such that $x_n \le x_{n+1}$ for all $n \in \mathbb{N}$.
Given that $A$ is bounded above, there exists $N \in \mathbb{R}$ such that $a \le N$ for all $a \in A$. Since, $x_n \in A$ for all $n \in \mathbb{N}$, it follows that $x_n \le N$. Thus, we've shown $(x_n)$ is bounded.
Finally, by the Monotone Convergence Theorem, we may conclude $\lim (x_n) = \sup \{ x_n : n\in \mathbb{N}\} = \sup A = u$. $\square$
I'm suspicious of the following thing regarding my proof. The open interval $(0,1)$ which is an infinite subset of $\mathbb{R}$ is uncountable, hence I cannot put it in one-one correspondence with the naturals. I've done that in my proof. Does the author mean a countable number of reals in the infinite subset of $\mathbb{R}$?
The problem is that $\min\{A\}$ doesn't necessarily exist (your $(0,1)$ example illustrates this). Even if it does, there is no reason your sequence covers all of $A$. Consider, for instance, $$A=\{1-1/n\mid n\in \Bbb N\}\cup (2,3),$$where your sequence only covers the lower part of the set: $x_n=1-\frac 1n$
The real way to do it is a bit less concrete. By definition of $\sup$, there is an $x_1\in A$ such that $u-1\leq x_1\leq u$ (it may be $u$ itself). Then there is an $x_2\in A$ such that $u-\frac12\leq x_2\leq u$. And so on.