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: 
This is not a direct answer to your question but it does complement other people's answers.

You need to be clear what statement is:

Bolzano-Weierstrass theorem says every bounded sequence $\{x\_n\}\_{n\in\mathbb{N}}$ in $\mathbb{R}^n$ has a convergent subsequence in $\mathbb{R}^n$.

Sequential compactness theorem says that every sequence $\{x\_n\}\_{n\in\mathbb{N}}$ in a closed and bounded subset of $\mathbb{R}^n$ has a convergent subsequence **in that set**.

Now it is not true that if $\{x\_n\}\_{n\in\mathbb{N}}$ is a sequence in $(0,1)$ then it has a convergent subsequence **in the open interval** $(0,1)$, as pointed out by Dominic. But it is true that $\{x\_n\}\_{n\in\mathbb{N}}$ has a convergent subsequence **in $\mathbb{R}$** as result of Bolzano-Weierstrass. 

So the Bolzano-Weierstrass theorem does not fail for open intervals, but the sequential compactness property does. Since $(0,1)$ is not closed, it does not contain all the limit points. If limit point that the subsequence of $\{x\_n\}\_{n\in\mathbb{N}}$ that wants to converge to is not in the open interval, like in the example, it does not satisfy sequential compactness property but Bolzano-Weierstrass is still correct.
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 $(0,1)$.