https://en.m.wikipedia.org/wiki/Least-upper-bound_property
Wiki provides a proof of least-upper-bound property of real numbers(set of equivalence classes of cauchy sequences over Q). proof is as follows;
Let $S$ be nonempty bounded subset. take any $a$ of $S$. name it A1. take any upper bound and name it B1. for n+1'th elements,
if An+Bn/2 is upper bound, then An+1=An and Bn+1=An+Bn/2.
if not, take An+1= s where s>An+Bn/2 and s belongs to S. Bn+1=Bn.
Wiki states that Bn-An goes to zero but I don't see clearly. Any hints will be thankful.
The simplest answer is using the monotone convergence theorem: If a sequence is bounded and monotone (only increases or decreases), it will be convergent. The sequence $b_n - a_n$ is bounded by $b_1 - a_1$ and $b_n - a_n \geq b_{n+1} - a_{n+1}$, hence the sequence is monotonically decreasing.
If you want to do it by definition, you need to observe that for each element, $$b_{n+1} - a_{n+1} \leq \frac{b_n - a_n}{2}.$$ Hence, $$b_n - a_n \leq \frac{b_1 - a_1}{2^{n-1}}.$$ Given a sequence and $\epsilon > 0$, pick an $N \in \mathbb{N}$ such that $\frac{b_1 - a_1}{2^{N-1}} \leq \epsilon$.