Why can't the Bolzano-Weierstrass theorem be proved for open intervals? 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!
 A: 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.
A: 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.
A: 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.
A: 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 $(0,1)$.
