Question: Please help me, I can't find the theorem anywhere that states something like:

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> For a bounded set $U\subset\mathbb R$ there exists a non-descreasing sequence $(a\_n)\_{n\in \mathbb N}$ with $a\_n \in U$ with $\lim\_{n\to\infty}a\_n=\sup U$.
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Thank you!

Answer: 
You can prove the theorem (under the assumption that $\sup U$ exists) directly from the definition of supremum. For each $n$ there is some point in $U$ within $1/n$ of the supremum. Use the axiom of choice to choose a sequence $(a\_n)$ so that for each $n$, $a\_n$ is within $1/n$ of the supremum. Then prove that this sequence converges to the supremum.
Because it follows directly from the definition of supremum. You can create the sequence $\{a\_n\}$ easily:
Let $M = \sup U $.
Assume that $M \notin U$ (otherwise just let $a\_i = M \, \forall i \in \mathbb{N}$).
Fix $a\_0 \in U$.
Choose $a\_i \in (\frac{M + a\_{i-1}}{2}, M) \cap U$ for $i = 1, 2, \dots$
Such $a\_i$ must exist since $M$ is the supremum. Otherwise $\frac{M + a\_{i-1}}{2} < M$ would be a lower upper bound for $U$.