$ \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 $I_{n}=[a_{n},b_{n}]$, where $a_{n}\leq b_{n}$. We go to prove
that $\sup_n a_{n}\leq\inf_n b_{n}$ by contradiction. Denote $a=\sup_{n}a_{n}$
and $b=\inf_n b_{n}$. Suppose the contrary that $a>b$. Choose $l\in(b,a)$.
Then there exists $n_{1}$ and $n_{2}$ such that $a_{n_{1}}>l$ and
$b_{n_{2}}<l$. Take $N=\max(n_{1},n_{2})$. By assumption, $\cap_{k=1}^{N}[a_{k},b_{k}]\neq\emptyset$,
so there exists $x_{0}\in\cap_{k=1}^{N}[a_{k},b_{k}]$. Now $x_{0}\in[a_{n_{2}},b_{n_{2}}]$
implies that $x_{0}\leq b_{n_{2}}<l$. On the other hand, $x_{0}\in[a_{n_{1}},b_{n_{1}}]$
implies that $x_{0}\geq a_{n_{1}}>l$. Contradiction!
Hence $a\leq b$. We assert that $[a,b]\subseteq\cap_{k=1}^{\infty}[a_{k},b_{k}]$
and it will follow that $\cap_{k=1}^{\infty}[a_{k},b_{k}]$ is non-empty.
Let $x\in[a,b]$. Let $k\in\mathbb{N}$ be arbitrary. Then $a_{k}\leq a\leq x\leq b\leq b_{k}$
implies that $x\in[a_{k},b_{k}]$. Q.E.D.
