Is every open interval a union of half open intervals?

If $M = (a,b)\cap \mathbb{Q}$ then $$(a,b) = \bigcup _{c\in M} [c,b)$$

$$(a,b) = \bigcup_{n=1}^\infty \, \left[\, \left(1-\frac{1}{n}\right)\, a + \frac{1}{n} b ,\, b\right) $$

$$(a,b) = \bigcup \{[x,b): a < x < b \}$$

Every $[x,b) \subseteq (a,b)$ whenever $a < x < b$ for the right to left inclusion, and on the other hand, if $a < x < b$, $x \in [x,b)$, which shows the left to right inclusion. If you want a countable union at all cost (topologies are closed under all unions, but maybe you're doing measure theory?) then take $x$ to be all rationals in $(a,b)$ so

$$(a,b) = \bigcup \{[q,b): q \in \mathbb{Q} \text{ and } a < q < b\}$$

but then the proof of equality is a little more involved: the right to left inclusion stays the same, but if $p \in (a,b)$ we first pick $q\ in \mathbb{Q}$ with $a < q < p$, and note that $p \in [q,b)$ which is a subset of the right hand side.