Bob and Alice are going to play a math game. The game has a countless number of rounds. In the nth round.
Boba chooses a family $ F_n $ dfe rational open ranges (ie: data $ p, q \in \mathbb{Q}$, the range is the set $\left\{x \in \mathbb{Q} | p <x <q \right \}$ such that
$$\mathbb{Q} = \bigcup_{I \in F_n } I$$.
We say that this family $F_n$ is an open cover of $\mathbb{Q}$.
Alice chooses an $I_n \in F_n$ range
Alice wins the game if $\bigcup_ {n=1} ^ {\infty} I_n = \mathbb{Q}$, and Bob wins if he prevents Alice from winning. Display a winning strategy for Alice.
Attempt: Let $r_1 < r_2 < r_3$ be a well ordering of the rational numbers. Then in round $n$, Alice should choose an interval that contains $r_n$. This will guarantee that she covers $\mathbb{Q}$ in a countably infinite number of rounds
I don't think it needs to be ordered, but why?
What if the families Bob chooses are not intervals but arbitrary unions or finite intersections of intervals?
It does have to be well-ordered, but that is a side property of what you need. Simply choosing a well-ordering of the rationals will not guarantee Alice a win. She needs an enumeration, as you have implicitly assumed and lulu has also mentioned.
By enumerating the rationals $\Bbb Q = \{r_i\}_{i\in \Bbb N}$, you have well-ordered it, namely with the order $r_i \prec r_j \iff i < j$. So a well-ordering in necessary.
But it is not sufficient. If Alice chooses an enumeration $\{q_i\}_{i\in \Bbb N}$ of $\Bbb Q\setminus \{0\}$ and similarly defines $q_i \prec q_j \iff i < j$, and then continues to define $q_i \prec 0$ for all $i$, this also provides a well-ordering of $\Bbb Q$, but one in which Alice can theoretically lose, as she and Bob are not required to make another pick after doing infinitely many, and thus she may not ever pick a set containing $0$.
As for your other question, it does not matter what form the sets take. All that is required is that at each step Bob's family covers all of $\Bbb Q$ so that Alice can pick a set holding her required rational.