$ \bigcap\limits_{n=1}^N I_n \neq \emptyset$ for all $ N \in \mathbb{N}$ implies that $ \bigcap\limits_{n=1}^\infty I_n \neq \emptyset $? How can I show that a sequence of closed bounded (Not necessarily nested) intervals $ I_1, I_2, I_3 ,\ldots$ with the property that $ \bigcap\limits_{n=1}^N I_n \neq \emptyset$  for all $ N \in \mathbb{N}$ implies that $ \bigcap\limits_{n=1}^\infty I_n \neq \emptyset $ ?
I'm being asked to determine if this is true.  I think that it is, because no matter how large $N$ is, we can always find an element in  $ \bigcap\limits_{n=1}^N I_n$.  So, we could simple let $N$ grow and there will always be an element in the intersection.  I was thinking about using induction, but this doesn't seem like an induction problem.  I am new to Real Analysis (self-study).  Someone tried to explain this to me using the bolzano weierstrass theorem, but I have not learned that. Any guidance will be appreciated.

Let $\displaystyle c_N = \inf \bigcap_{n=1}^N [a_n,b_n] $ and $\displaystyle d_N= \sup \bigcap_{n=1}^N [a_n,b_n].$
See if you can show that $c_1\le c_2\le c_3\le \cdots \le c_N \le \cdots \le d_M \le \cdots \le d_3\le d_2\le d_1$ for all $N,M.$
Then see if you can show that $\displaystyle \varnothing\ne \left[ 
\sup_n c_n, \inf_n d_n \right] \subseteq\bigcap_n [a_n,b_n].$
