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.

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 $\bigcap\limits_{i=1}^nC_i=\emptyset$.
