Let $I_n = [a_n,b_n]$ be a nested family of closed intervals, that is, $a_n,b_n \in \Bbb R$, $a_n < b_n$, with $a_n ≤ a_n+1$ and $b_n ≥ b_n+1$ for every $n$. Assume that $\lim_{n\to\infty}(b_n − a_n) = 0$. Suppose $\{x_n\}_{n≥1}$ is a sequence of real numbers, where $x_n \in I_n$ for every $n$. Show that $\{x_n\}_{n≥1}$ is convergent.
As the sequence $(a_n)$ and $(b_n)$ are both convergent and converges to the same limit hence the nested interval has a convergent sequence now $(x_n)$ belongs to $(a_n)$ or $(b_n)$ or intersection of both two. Now if it belongs to $a_n$ or $b_n$ then convergent but what if it belongs to its intersection As the intersection will be non empty then how to proceed?
Suppose that $(x_n)_n$ does not converge.
Then it is not a Cauchy-sequence so an $\epsilon>0$ exists such that for all $n\in\mathbb N$ we can find $k,m\geq n$ with $|x_k-x_m|>\epsilon$.
However, we have $a_n\leq x_k\leq b_n$ and $a_n\leq x_m\leq b_n$ so that $|x_k-x_m|\leq b_n-a_n$.
This leads to a contradiction.
Do you see how?