Why does the following proof of the Nested Interval Property require the Axiom of Completeness?

The proof in my text is as such:

Let $a_1,a_2,a_3,\ldots$ and $b_1, b_2, b_3,\ldots$ be the labels of the left- and right- hand endpoints respectively.

Consider the set A of the left-hand endpoints of the intervals, and let x = sup A. Since x is an upper bound for A, we have $a_n \leq x$. Since each $b_n$ is an upper bound for A, we have $x\leq b_n$. Then since $a_n\leq x \leq b_n$, we can conclude that $x\in I_n$ for every choice of $n\in \mathbb{N}$. Hence x is in the infinite intersection of nested intervals.

My question is.. could the proof work with $x = a_n$ instead? It seems that all the key properties would still hold — $a_n \leq a_n \leq b_n$ for all n. This does not seem to require the AoC to be true.

Solutions Collecting From Web of "Why does the following proof of the Nested Interval Property require the Axiom of Completeness?"

Take some irrational number, e.g. $\sqrt{2}$. Consider its decimal expansion
$$1.4142\ldots$$
Now construct a sequence of nested intervals:
$$[1,2] \supset [1.4,1.5] \supset [1.41,1.42] \supset [1.414,1.415] \supset [1.4142,1.4143] \supset \cdots$$
These are nested intervals with rational endpoints, but their intersection contains no rational number. So without completeness the Nested Interval Property is false.