Euclidean space Question: Let $\{F_n\mid n \in \mathbb{N}\}$ be a family of non empty closed subsets of the Euclidean space $\mathbb{R}^p$ with $F_1$ bounded and $F_{n+1} \subset F_n$ for every $n\in\mathbb{N}$. Show that $\bigcap_{n\in\mathbb{N}}F_n\neq\emptyset$.
Source: It was asked in an exam in the year 2005.
 A: Let's rephrase this question as:

Let $F_1$ be a closed, bounded subset of $\mathbb R^p$ and let $F_n$ be a sequence of closed nonempty subsets of $F_1$ such that $F_{n+1}\subset F_n$ for all $n\in\mathbb N$. Show that $\cap_{n\in\mathbb N} F_n\ne \emptyset$.

This is the same as saying that $F_1$ satisfies what's called the finite-intersection property. The crucial insight for this problem is that the finite-intersection property is equivalent to compactness. I'll prove the direction you need.
Theorem: Let $F$ be a compact space and $F_n$ a sequence of closed nonempty subsets such that $F_{n+1}\subset F_n$ for all $n\in\mathbb N$. Then $\cap_{n\in\mathbb N}F_n\ne\emptyset$.
Proof: Suppose otherwise. Then $\{F\setminus F_n\}_{n\in\mathbb N}$ is an open cover, so we have some $N$ such that $\{F\setminus F_n\}_{n\leq N}$ is a subcover. But then $F\setminus F_N=F$, so $F_N=\emptyset$, a contradiction.
Now simply observe that $F_1$ is compact by Heine-Borel.
A: Here's a proof which uses the Bolzano-Weierstrass theorem.  Choose a point $x_i \in F_i$.   The sequence $\{x_i\}_{i = 1}^\infty$ lies in the sequentially compact set $F_1$, and therefore has convergent subsequence $\{x_{i_k}\}_{k=1}^\infty$ which converges to some $x \in F_1$.  For any given $n$, the subsequence lies in the closed set $F_n$ when $i_k \ge n$, and therefore $x \in \bigcap_{n \in \mathbb{N}} F_n$
