Proving $\bigcap_n I_n \neq \emptyset$ for monotonically smaller closed intervals $(I_n)_n$ Theorem: Let $(I_n)_n$ be monotonically smaller closed intervals. i.e. $I_{n+1}\subseteq I_n$
Then $$\bigcap_n I_n \neq \emptyset$$
I easily grasped this geometrically, But I doubt of my proof. This is my proof:

Since $I_n$'s are real closed intervals, $I_n \neq \emptyset$. Then we can take $x \in I_n$. Since $I_n\subseteq I_{n-1}$, $x \in I_{n-1}$.
And generally from that,
$$x \in I_{n-2} \Rightarrow x \in I_{n-3} \Rightarrow x \in I_{n-4} \Rightarrow...\Rightarrow x \in I_2 \Rightarrow x \in I_1$$
Thus $x \in I_{n}$ is also in other closed intervals $I_i$s where $i < n-1$. Thus,
$$\bigcap_{i \leq n} I_i \neq \emptyset$$ Then, if we just take $\lim_{n\to \infty}$ equation will still hold and consequently,
$$\bigcap_{n} I_n \neq \emptyset$$

To me, it looks right, but I could'nt be sure. I would appreciate your short-time look. Thanks from now.
 A: 
Then, if we just take $\lim_{n\to \infty}$ equation will still hold
and consequently, $$\bigcap_{n} I_n \neq \emptyset$$

This is bad reasoning. Just because some relation holds for all values of $n$, that does not mean the relation holds for the limit.
For example, the relation $\frac1n > 0$ holds for all $n\in\mathbb N$, but the relation $$\lim_{n\to\infty} \frac 1n > 0$$
does not hold.

Another way to see your proof is not OK is this:
If $I_n$ are open intervals, then your "proof" can be used to prove that $$\bigcap_n I_n \neq \emptyset$$
even though it is possbile to find a sequence of intervals such that their intersection is empty (for example, $I_n = (0, 10^{-n})$).

To improve your proof: well, it's not really a case of improving it, I'm afraid your best bet is to go back to the drawing board. Try to find a value $x$ that is in all intervals $I_n$. To find it, think about $I_n=[a_n, b_n]$, and then think about the sequences $a_1, a_2,\dots$ and $[b_1, b_2, \dots]$. What are some of their interesting properties?
A: For the case $I_n=[0,1/n]$ the number $1/100$
belongs to $I_1,I_2,\ldots,I_{100}$ but not to $I_{101}$
Hint: Use Bolzano-Weierstrass.
