Why doesn't the nested interval theorem hold for open intervals?

Why is the condition that the intervals be closed necessary? Could someone give me an example of a sequence of nonempty, bounded, nested intervals whose intersection is empty? I can't think of one, so why does the theorem require it?

Here is the theorem:

If $I_1 \supset I_2 \supset I_3 \supset \dots$ is a sequence of nested, closed, bounded, nonempty intervals, then $\bigcap_{n=1}^{\infty}I_n$ is nonempty. If, in addition, the length of $I_n$ approaches zero, then $\bigcap_{n=1}^{\infty}I_n$ consists of a single point.

My apologies if this has been asked before but I couldn't find it.

Consider the family $A_n = (0, \frac{1}{n})$. We have $A_{n+1} \subset A_n$ for every $n \in \mathbb{N}$, and the length of $A_n$ approaches zero as $n$ approaches infinity, but $\bigcap_{n=1}^{\infty} A_n$ is empty. To see this, note that any element of the intersection would be greater than zero, yet less than $\frac{1}{n}$ for every $n \in \mathbb{N}$; by the Archimedian property of the real numbers, there is no such element.

One way to look at it is to see why the proof of the NIT fails if you try open sets - the NIT proof shows that the infinite intersection becomes the singleton closed interval $[a,a] = \{a\}$. If you try it with open sets, you'll arrive at $(a,a) = \{\}$.