Sequence converging to the supremum

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.

Let $$M:=\left\{a\in\mathbb R:a\ge x\text{ for all }x\in U\right\}\;.$$ By definition, $$s:=\sup U=\min M\;.$$ Let $n\in\mathbb N$ $\Rightarrow$ $s-1/n<s$ and hence $$s-\frac1n\not\in M\;,\tag1$$ i.e. $\exists x_n\in U$ with $$s-\frac1n<x_n\le s\tag2\;.$$ The squeeze theorem yields $$x_n\xrightarrow{n\to\infty}s\tag3\;.$$ Now, let $$y_n:=\max(x_1,\ldots,x_n)\;\;\;\text{for }n\in\mathbb N\;.$$ Note that $(y_n)_{n\in\mathbb N}\subseteq U$ is nondecreasing with $$y_n\xrightarrow{n\to\infty}\lim_{n\to\infty}x_n=s\tag4\;.$$

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 $U$.