I need to prove that every bounded open interval in $\displaystyle \mathbb{R}$ is a union of disjoint open sets. Not sure, but here's what I got:

Let $\displaystyle (a,b)$ be an open interval and let $\displaystyle x,y \in (a,b)$ with $\displaystyle x \not = y$ and assume $\displaystyle x < y$. Put $\displaystyle \epsilon = \min\{d(x,y), d(x,a), d(y,b)\}$ and define $\displaystyle S = \cup \{ B(x;\epsilon) : x \in (a,b)\}$. Then $\displaystyle S=(a,b)$ and it is a union of open sets.

Now we need to show they are pairwise disjoint. Using the previous definition of $\displaystyle \epsilon$, there is an open ball $\displaystyle B(x; \epsilon /3)$ and the ball $\displaystyle B(y; \epsilon /3)$ with $\displaystyle B(x; \epsilon /3), B(y; \epsilon /3) \subset (a,b) $. Suppose some $\displaystyle z \in B(x; \epsilon / 3) \cap B(y; \epsilon /3)$. Then $\displaystyle \epsilon \leq d(x,y) \leq d(x,z) + d(z,y) \leq \frac {\epsilon} {3} + \frac {\epsilon} {3} = \frac{2\epsilon}{3}$ gives a contradiction.