Proving the intersection of nested sets $C_i$ cannot be empty unless there exists a $k$ such that $C_i = \emptyset$ $\forall i \geq k$ I'm working on a problem in my functional analysis course and am confused as to why the answer seems more convoluted than needs be. 
The problem is 

Prove that in a finite dimensional normed space, $X$, any decreasing sequence ($C_{i+1} \subseteq C_i$  $\forall i$ ) of closed, bounded sets $\{ C_i \} $ cannot verify $ \bigcap_{i \in \mathbb{N} } C_i = \emptyset $ unless there exists $k > 0$ such that $C_i = \emptyset$ $\forall i \geq k$. 

Summarising the lecturer's solution: After considering the case where $C_{i+1} = C_i$, a sequence is formed from $x_i \in C_i \backslash C_{i+1}$ which has a convergent subsequence converging to $x$, and by contradiction with the assumption that $x \notin   \bigcap_i C_i$ it is shown that $ x\in  \bigcap_i C_i$.
When I was trying to prove it I simply wrote that as $ C_{i+1} \subseteq C_i$ we have that $C_i \cap C_{i+1} = C_{i+1}$ thus $C_1 \cap \dots \cap C_{i+1} = C_{i+1} $. Then $\bigcap_{i \in \mathbb{N}} C_{i} = \emptyset $ iff $C_{k} = \emptyset$ for some $k$ and it follows $C_{j} = \emptyset$ $\forall j \geq k $.
I have a couple of questions, firstly, can someone clarify how showing a sequence in $C_i \backslash C_{i+1}$ converging to a value in $\bigcap_i C_i$ proves the question (the proof ends with the contradiction so I assume it is meant to be trivial).
Secondly, have I made any glaringly obvious mistakes in my approach, or perhaps over simplified and ignored a property?
I would appreciate some help and would be happy to add any extra details if needs be. Thanks in advance.
 A: Edit: of course, this can be done easily with sequences and Bolzano-Weierstrass. Note that your lecturer's argument seems to make a useless case distinction which does not simplify anything. If $C_k$ is not eventually empty, it is a fortiori nonempty for every $k$. Pick an element in each $C_k$. This sequence is bounded in a finite-dimensional normed vector space. By Bolzano-Weierstrass, it has a converging subsequence. Then the limit belongs to each $\overline{C_k}=C_k$. So the intersection of the $C_k$'s is not empty, and you get the contrapositive of what you want. Now the following argument works in an arbitrary toplogical space, provided $C_0$ is compact, and each $C_i$ is closed.
It is not "trivial". First note that in a topological space $X$, the following are equivalent:
1 - Whenever $X=\bigcup_{i\in I} U_i$ for an arbitrary family of open sets $U_i$ in $X$, there exists a finite subfamily such that $X=\bigcup_{j=1}^nU_{i_j}$.
2 - If $(F_i)_{i\in I}$ is an arbitrary family of closed sets in $X$ and if $\bigcap_{i\in I}F_i=\emptyset$, then there exists a finite subfamily such that $\bigcap_{j=1}^nF_{i_j}=\emptyset$.
Just take the complements to pass from one to the other one. By definition, $X$ is compact if it satisfies these equivalent conditions. Note that lots of people also require $X$ to be Hausdorff. But this has no incidence here as we are in a normed vector space.
Now in a finite-dimensional normed vector space, compact=closed+bounded (Heine-Borel). In particular, $C_0$ is compact. Note that every $C_j$ is closed in $C_0$. If $\bigcap_j C_j=\emptyset$, property 2 above yields the existence of a finite subfamily such that $\bigcap_{j=1}^nC_{i_j}=\emptyset$. Now since the sequence $C_j$ is nonincreasing
$$
\bigcap_{j=1}^nC_{i_j}=C_{\max i_j}=C_{k_0}=\emptyset
$$
hence $C_k\subseteq C_{k_0}=\emptyset$ implies $C_k=\emptyset$ for all $k\geq k_0$.
Note: if you simply assume that the $C_i$'s are closed, this is no longer true, as proven by the example $C_i=[i,+\infty)$ in $\mathbb{R}$. A related, true, fact is the following: if $F_n$ is a nonincreasing sequence of nonempty closed sets in a complete space, and if the diameters of the $F_n$'s tend to $0$, then $\bigcap_{n}F_n$ is nonempty, and equal to a singleton.
A: Suppose that $C_{i+1}\subseteq C_i$ and for all $n$, $\bigcap\limits_{i=1}^nC_i\ne\emptyset$. Thus, for all $n$, we can find an $x_n\in\bigcap\limits_{i=1}^nC_i$. Since each $x_n\in C_1$, and $C_1$ is closed and bounded, there is a subsequence $x_{n_k}$ that converges to $x_\infty$.

For each $i$, $C_i$ is closed and for all $n_k\ge i$, $x_{n_k}\in C_i$. Therefore, $x_\infty\in C_i$ for each $i$. That is,
$$
x_\infty\in\bigcap_{i=1}^\infty C_i
$$
Thus, assuming that  for all $n$, $\bigcap\limits_{i=1}^nC_i\ne\emptyset$, we get that $\bigcap\limits_{i=1}^\infty C_i\ne\emptyset$. The contrapositive says that if $\bigcap\limits_{i=1}^\infty C_i=\emptyset$, then there is an $n$ such that $\bigcap\limits_{i=1}^nC_i=\emptyset$.
