Functional Analysis proof help involving decreasing sequence of sets in a finite dimensional normed space Hi I am struggling with a proof;
Prove that in a finite dimensional normed space, Z, any decreasing sequence of closed and bounded sets $\{C_i\}$ ($C_{i+1}\subseteq C_i\;\forall i$) cannot satisfy $\displaystyle\cap_{i\in\mathbb{N}}C_i=\emptyset$ unless there exists $k>0$ such that $C_i=\emptyset\;\forall i\geq k$.
Now I attempted an initial solution and looked at the first couple of steps of the answer and saw I was completely wrong so I didn't look at the rest of the answer and I really want to do as much of it myself as I can so some pushes in the right direction would be appreciated.
Now considering our sequence $\{C_i\}$ if $\exists j\in\mathbb{N}$ st $C_j=C_{j+n}\;\forall n>0$ the statement follows immediately so suppose not. Then WLOG we may assume that $C_{i+1}\subsetneq C_i\;\forall i$. 
Define $X_i=C_i - C_{i+1}\neq \emptyset$ and construct a sequence $\{x_i\}$ by selecting $x_i \in X_i$. Now I'm not really sure where to go, I know that since $\{x_i\}$ is contained in C_1 and the set is closed and bounded the sequence is bounded and hence contains a convergent subsequence which converges to a point in C_1 but I don't know where to go next
 A: Suppose not, i.e. there is a subsequence $n_1<n_2<\cdots$ such that $C_{n_k}\neq\varnothing$ for all $n_k$. Then there is a sequence $x_{n_k}$ with $x_{n_k}\in C_{n_k}$. If $\dim Z = d$ then $Z$ is isomorphic to $\Bbb R^d$, so by the Heine-Borel property, each $C_n$ is compact. As $x_{n_k}$ is a sequence in a compact metric space, it has a convergent subsequence, again denoted $x_{n_k}$, with its limit in $C_{n_1}$. Applying this argument to the sequence $x_{n_2},x_{n_3},\dots$ we see that this subsequence has a limit in $C_{n_2}$. So for any $n_k$, we may find a convergent subsequence with a limit in $C_{n_k}$. Denote these limits $y_k$. Since $\bigcap_{k=1}^\infty C_{n_k}=\varnothing$, we must have $\sup\{\|x-y\| : x,y\in C_{n_k}\}\to 0$ as $k\to\infty$. It follows that $y_k$ is a Cauchy sequence, and as $Z$ is complete, it has a limit $y$. Since for any $\varepsilon>0$ we can find $N$ such that $n_k\geqslant N$ implies $\|y_k-y\|<\varepsilon$, we can find infinitely many $n_k$ such that $y\in C_{n_k}$. Hence $y\in\bigcap_{k=1}^\infty C_{n_k}=\bigcap_{n=1}^\infty C_n$, a contradiction.
(Note: There's probably a more elegant way to do this and there could be an error or two in my proof so let me know if you see anything wrong).
