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.
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.