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: 
Hint: Think about the sequence $$\left\{\frac{1}{n}\right\}\_{n=2}^\infty\subset(0,1).$$The problem is that the bisection may put you closer and closer to an endpoint, which isn't contained in an open interval.
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 $(0,1)$.