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My proof is similar in spirit to this one, but I'm not able to figure out if it's correct or not.

Consider $S\subset \mathbb{R}$ that is bounded above. Consider an upper bound of $S$, say $u$. If $u\in S$, we are done, so let $u\notin S$. We will construct $I_n = [a_n,b_n]$ to apply the nested-interval theorem. Define $b_i = u$, for all $i$.

For $a_1$, pick some $s\in S$. Now define $a_n$ recursively (is it okay to make recursive definitions like this one):

  1. If there exists $s'\in S$, such that $s' > a_i$ put $s_{i+1} = s'$.
  2. Otherwise, $s_{i+1} = s_i$.

We can see that $$I_1\supseteq I_2 \supseteq ...$$ By the nested interval theorem $$\bigcap_{n\ge 1} I_n = [a,b] \neq \varnothing$$ We can now show that $a$ is the least upper bound of $S$.

Is this proof alright?

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I think your definition of $a_n$ is a problem. Suppose $S = [0,2]$. This is clearly bounded. Pick $u=3$ and the we can pick $a_n=1-\frac{1}{n}$. We then get a nested series of closed intervals but their intersection is $[1,3]$. $1$ is clearly not an upper bound for $S$.

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    $\begingroup$ Right! Any ideas how to fix this? $\endgroup$
    – stoic-santiago
    Jan 18, 2021 at 7:18
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    $\begingroup$ I would try something like this, that controls the length of the intervals $I_n$: Pick $a_1$ some element in $S$, $b_1$ an upper bound, let $I_1=[a_1,b_1]$. Then let $m_1 = \frac{a_1+b_1}{2}$. If $[m_1,b_1]$ contains no element of $S$, pick $I_2=[a_1,m_1]$. Otherwise, pick $I_2=[m_1,b_1]$. Rename the endpoints as $I_2=[a_2,b_2]$ and proceed recursively. $\endgroup$
    – Anonymath
    Jan 18, 2021 at 7:25

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