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.

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.
