Please bear with me. The question might look huge but I just included the relevant definitions for making the post self-contained:
My definition of atlas:
Let $M$ be a set. We call a set $\mathcal{A} = \{(U_i, \phi_i)\mid i \in I\}$ of local charts an atlas of dimension $m$ on $M$ if the following conditions are satisfied:
(1) $U_i \subseteq M$
(2) $\phi_i: U_i \to \mathbb{R}^m$ is injective.
(3) $\phi(U_i)$ is open.
(4) $\bigcup_{i \in I} U_i = M$
(5) If $i,j \in I, U_i \cap U_j \ne \emptyset$, then $\phi_i(U_i \cap U_j)$ is open.
(6) If $i,j \in I, U_i \cap U_j \ne \emptyset$. Then $\phi_j \circ \phi_i^{-1}: \phi_i(U_i \cap U_j) \to \phi_j(U_i \cap U_j)$ is smooth (of class $C^\infty$).
Definition of canonic topology:
Given an atlas $\mathcal{A}$ of $M$, we define the canonic topology on $(M, \mathcal{A})$ as the set of all unions of domains of local charts equivalent with $\mathcal{A}$. This topology does not change if we replace $\mathcal{A}$ by an equivalent atlas.
I want to prove:
Let $M$ be a set with atlas $\mathcal{A} = \{(U_a, \phi_a)\}_{a \in A}$. Then the canonic topology on $(M, \mathcal{A})$ is given by $\mathcal{O}:= \{V \subseteq M \mid \forall a \in A: \phi_a(V \cap U_a) \mathrm{\ open}\}$
My attempt:
Write $\mathcal{T}$ for the canonic topology. We prove that $\mathcal{T} \subseteq \mathcal{O}$.
Let $V \in \mathcal{T}$. Then $V = \bigcup_{i \in I} O_i$ for local charts $(O_i, \psi_i)$ equivalent with $\mathcal{A}$. It is easy to see that $\mathcal{O}$ is closed under unions, so it suffices to show that $O_i \in \mathcal{O}$.
For this, we need to show that $\phi_a(O_i \cap U_a)$ is open for every $a \in A, i \in I$.
Fix $i \in I$. By compability of the chart $(O_i, \psi_i)$, we have that $\mathcal{A} \cup \{(O_i, \psi_i)\}$ is an atlas, and for every $a \in A$, $\phi_a(O_i \cap U_a)$ is open, by definition of atlas (more specifically, see (5) in my definition)
For the other inclusion, let $V \in \mathcal{O}$. Then $\phi_a(V \cap U_a)$ is open for every $a \in A$ and
$$V= V \cap M = V \cap (\bigcup_{a \in A} U_a) = \bigcup_{a \in A} (V \cap U_a)$$
so it suffices to show that there are local charts with domain $V \cap U_a$ compatible with $\mathcal{A}$.
For this, consider the chart $(U_a \cap V, \phi_a\vert_{U_a \cap V})$. Because $V \in \mathcal{O}$, we have that $\phi_a(U_a \cap T)$ is open, so this is a local chart of $M$.
It remains to check that it is compatible with $\mathcal{A}$. But
$\phi_a(U_a \cap V \cap U_b) = \phi_a(U_a \cap V) \cap \phi_a( U_a \cap U_b)$ is open, as intersection of open sets. Similarly, $\phi_b(U_a \cap V \cap U_b)$ is open and it is also straightforward to check that the transitions between the local charts are $C^\infty$. Hence, $(U_a \cap V, \phi_a\vert_{U_a \cap V})$ is compatible with $\mathcal{A}$ for all $a \in A$ and we are done.
Questions:
(1) Is this proof correct?
(2) I essentially gave the proof my textbook provided, but there they proved that $\mathcal{O}$ is a topology and that $\mathcal{O}$ does not depend on the chosen atlas. Why is this necessary to prove this?
(1) Yes, it's entirely correct, good work.
(2) It's never necessary to do anything in life. But for example, proving first that $\mathcal{O} = \mathcal{T}$ saves the trouble of proving that $\mathcal{T}$ itself is a topology, if you show that $\mathcal{O}$ is a topology, which seems a bit easier. On the other hand, it's easier to argue that $\mathcal{T}$ does not depend on the atlas than $\mathcal{O}$. So it's convenient to have both characterizations of that topology, I guess.