Let $\{E_n\}$ be a sequence of closed bounded sets in a complete metric space $X$, such that $E_{n+1} \subset E_n$, for all $n \in \mathbb{N}$, and $\displaystyle \lim_{n \to \infty} $diam $E_n=0$. Prove that $\bigcap_{n=1}^{\infty} E_n$ contains exactly one point.
Proof:
Let $E= \bigcap_{n=1}^{\infty} E_n$; by definition, for any $n \in \mathbb{N}$, $E \subset E_n$ and so diam $E \leq$ diam$E_n$. Since, diam $E_n \to 0$, assuming $\lvert E \rvert > 1$ leads to a contradiction. So, the proof reduces to showing that $E$ is not empty.
Edit: Note that we assume the $E_n$ are all non-empty. Further, if any $\lvert E_k \rvert =1$, the result follows immediately, so, for all $E_n$, we assume $\lvert E_n \rvert > 1$.
For any $E_n$, there exists $q_n \in E_n$ and a smallest real number $M_n>0$, such that $$d(p, q_n) \leq M_n, \qquad \forall p \in E_n.$$
Since we must have $M_n \to 0$, for any $\epsilon >0$, there exists an integer $N$ such that
$$d(q_n, q_m) \leq d(q_n, p) + d(p, q_m)< M_N + M_N< \epsilon,$$
for all $n \geq N, m \geq N$ (by an appropriate choice of $p$).
The sequence of points $\{q_n\}$ is thus a Cauchy sequence in $X$, and so it converges to some point $q \in X$.
For some $K \in \mathbb{N}$, assume $q$ is not a point in $E_K$, then $q \notin E_n$ for all $n \geq K$. It also follows that $q$ is not a limit point for $E_K$, and so there exists $\epsilon>0$, such that
$$p \notin N_\epsilon (q), \quad \forall p\in E_K.$$
The convergence of $\{q_n\}$ implies that there exists an integer $m > K $ such that
$$d(q_m, q)< \epsilon,$$
but
$$E_m \subset E_K \implies q_m \in E_K,$$
which is a contradiction. Thus $q \in E$. $\qquad \square$
Question:
Is there an additional assumption in the problem that the $E_n$ are all non-empty (since the empty set is closed, bounded, and is its own subset)?
Best Answer
The assumption that the $E_n$ are non-empty might be hidden in the assumption about their diameter, depending on your definitions. Usually if $E$ is a subset of a metric space $(M,d)$, we denote:
$$ \operatorname{diam} E = \sup \{ d(x,y) : x, y \in E \} $$
If, however, $E$ is empty, then we are taking the supremum of the empty set. This is usually defined to be $-\infty$. So, if one of the $E_n$ is empty, then all of the following sets are too, which means that the limit of their diameters would be $-\infty$: a contradiction to the assumption.
However, this does rely on this particular definition of the supremum. Usually, it's best to be clear and explicitly state the $E_n$ are non-empty.