Question: The proof I was given used interval bisection and if I apply that proof to an open interval I don't see where it would go wrong. 

Also is it true that the Bolzano Weierstrass property doesn't hold for any open interval? 

Thanks!

Answer: 
the Bolzano proof uses the lemma that a nested sequence of closed intervals has non-empty intersection.

The same counter-example others have given is esssentially the counter-example for this lemma.

Namely:
$$I\_n =(0,1/n)$$

is a nested sequence of open intervals whose intersection is empty.
Because the Intersection would be empty, if you do the same.  

When I think of the same proof you do, we need the fact that if
$$C\_{i+1}\subset C\_{i}$$ for all $i$, 
And $C\_i$ are closed intervalls 
$$\operatorname{diam} C\_{i+1} \leq \frac{1}{2} \operatorname{diam} C\_i$$
where diam is the length of the intervall, then
$$\bigcap\_{i=1}^\infty C\_i=\{ x\_0 \}$$
For open Intervalls this one is not true take
$$\bigcap\_{i=1}^\infty \left(0,\frac{1}{2^i}\right) =\varnothing$$

For example taking the open Interval $(0,1)$ and the sequence 
$$a\_n=\frac{1}{n}$$
this one is bounded, but it has no limit point in $(0,1)$.