There are two consequences to nesting of intervals: $[a_{m},\,b_{m}]\subseteq[a_{n},\,b_{n}]$ for $n\leq m$:

1. 1.

   first of all, we have the inequality $a_{n}\leq a_{m}$ for $n\leq m$, which means that the sequence $a_{1},a_{2},\ldots,a_{n},\ldots$ is nondecreasing;

2. 2.

   in addition, we also have two inequalities: $a_{m}\leq b_{n}$ and $a_{n}\leq b_{m}$. In either case, we have that $a_{i}\leq b_{j}$ 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 $(a_{i})$, $a\leq b_{i}$ for all $i$. But because $b$ is the infimum of $(b_{i})$, $a\leq b$. Therefore, the interval $[a,b]$ is non-empty (containing at least one of $a,b$). Since $a_{i}\leq a\leq b\leq b_{i}$, 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 $a$. ∎