Proposition 1   If $$[a_1,\,b_1]\;\supseteq [a_2,\,b_2]\;\supseteq [a_3,\,b_3]\;\supseteq\ldots$$ is a sequence of nested closed intervals, then
If also $\displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$ , then the infinite intersection consists of a unique real number.

Proof. There are two consequences to nesting of intervals: $[a_m,\,b_m]\subseteq[a_n,\,b_n]$ for $n\le m$ :

1. first of all, we have the inequality $a_n\le a_m$ for $n\le m$ , which means that the sequence $a_1, a_2, \ldots, a_n, \ldots$ is nondecreasing;
2. in addition, we also have two inequalities: $a_m\le b_n$ and $a_n\le b_m$ . In either case, we have that $a_i\le b_j$ for all $i,j$ . This means that the sequence $a_1, a_2, \ldots, a_n, \ldots$ is bounded from above by all $b_i$ , where $i=1,2,\ldots$ .

Therefore, the limit of the sequence $(a_i)$ exists, and is just the supremum, say $a$ (see proof here). Similarly the sequence $(b_i)$ is nonincreasing and bounded from below by all $a_i$ , where $i=1,2,\ldots$ , and hence has an infimum $b$ .

Now, as the supremum of $(a_i)$ , $a\le b_i$ for all $i$ . But because $b$ is the infimum of $(b_i)$ , $a\le b$ . Therefore, the interval $[a,b]$ is non-empty (containing at least one of $a,b$ ). Since $a_i\le a\le b\le b_i$ , every interval $[a_i,b_i]$ contains the interval $[a,b]$ , so their intersection also contains $[a,b]$ , hence is non-empty.

If $c$ is a point outside of $[a,b]$ , say $c<a$ , then there is some $a_i$ , such that $c<a_i$ (by the definition of the supremum $a$ ), and hence $c\notin [a_i,b_i]$ . This shows that the intersection actually coincides with $[a,b]$ .

Now, since $\displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$ , we have that $b-a=\displaystyle\lim_{n\to\infty}b_n - \displaystyle\lim_{n\to\infty} a_n = \displaystyle\lim_{n\to\infty}(b_n-a_n) = 0$ . So $a=b$ . This means that the intersection of the nested intervals contains a single point $a$ .