Open set as a countable union of open bounded intervals

Hint: Let $A$ be open, and consider all intervals $(p,q)$ such that $p$ and $q$ are rational and $(p,q)\subset A$.

Yes. The answers to this question show that every non-empty open subset of $\Bbb R$ can be written as the union of countably many pairwise disjoint open intervals, when sets of the form $(a,\to)$ and $(\leftarrow,a)$ and $\Bbb R$ itself are counted as intervals. To complete the argument, you need only show that these open rays can be written as the union of countably many bounded open intervals:

$$\begin{align*} (a,\to)&=\bigcup_{n\in\Bbb Z^+}(a,a+n)\\\\ (\leftarrow,a)&=\bigcup_{n\in\Bbb Z^+}(a-n,a)\\\\ \Bbb R&=\bigcup_{n\in\Bbb Z^+}(-n,n) \end{align*}$$

Added: You can’t get pairwise disjointness in these cases, but you can arrange matters so that every point is in at most two of the bounded intervals, indeed, in the closures of at most two of them:

$$\begin{align*} (a,\to)&=\bigcup_{n\in\Bbb N}(a+2n,a+2n+3)\\\\ (\leftarrow,a)&=\bigcup_{n\in\Bbb Z^+}(a-2n-3,a-2n)\\\\ \Bbb R&=\bigcup_{n\in\Bbb Z^+}\Big((2n,2n+3)\cup(-2n-2,-2n+1)\Big)\;, \end{align*}$$

where $\Bbb N$ is the set of non-negative integers (i.e., it includes $0$).

If you don’t care about how much overlap your intervals have, of course, you can simply use the fact that the open intervals with rational endpoints are a countable base for the topology of $\Bbb R$.