$ \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 will reduce the problem to the The Nested Interval Theorem (also known as Cantor's intersection theorem). Let's look at the following sets $$A_n=\bigcap\limits_{k=1}^{n}I_k$$ Each $A_n$ has the following properties

• $A_n \ne \varnothing $, this is given
• $A_n$ is closed
• $A_n$ is bounded

And $$A_{n+1}=\bigcap\limits_{k=1}^{n+1}I_k=\left(\bigcap\limits_{k=1}^{n}I_k\right)\bigcap I_{n+1}=A_n\bigcap I_{n+1}$$ or $A_{n+1} \subset A_{n}$ (because $\forall x \in A_{n+1} \Rightarrow x \in A_{n}$). According to The Nested Interval Theorem $$\bigcap\limits_{k=1}^{\infty}A_k \ne \varnothing$$ But $$\bigcap\limits_{k=1}^{\infty}A_k=\bigcap\limits_{k=1}^{\infty}I_k$$