If $ A_1, A_2,...$ are sets with volume and $A= \cup_{i=1}^\infty A_i$ is a bounded set, must $A$ have volume?
This was a homework problem that we went over in class, and if I remember correctly the answer was: no. However, I'm having some trouble coming up with a counter example. I think the idea is to have it so that $\partial A$ does not have measure zero, and $A = \mathbb {Q} \cap [0,1]$ comes to mind, but then each of the $A_i$ may not have volume.
From context it sounds like you are talking about Jordan measure. In this case the answer to your question is no; this is one of the major reasons analysts tend to prefer the Lebesgue measure over the Jordan measure.
You are already on the right track for building a counterexample. Consider $\{ q_i \}_{i=1}^\infty$ an enumeration of $A=\mathbb{Q} \cap [0,1]$, then consider $A_i=\{ q_i \}$. Now $\mathbb{Q} \cap [0,1]$ is not Jordan measurable, which you could prove by noting that $1_{\mathbb{Q} \cap [0,1]}$ is nowhere continuous. By contrast $A_i$ are Jordan measurable, which you could prove by noting that $1_{A_i}$ is continuous except at one point. Other proofs are available; which one is convenient for you depends on how your definitions are stated and what theorems you have already established.