Reading through Tu's an introduction to manifolds, where some topological notions are given in chapter 2, section 7.1. More specifically "quotient topology" is briefly explained.
I quote the relevant bits first:
The equivalence class $[x]$ of $x \in S$ is the set of all elements in $S$ equivalent to $x$ An equivalence relation on $S$ partions $S$ into disjoint equivalence classes. We denote the set of equivalence classes by $S / \sim$ and call this set the quotient of $S$ by the equivalence relation $\sim$. There's a natural projection map $\pi : S \to S / \sim$ that sends $x \in S$ to its equivalence class $[x]$
First silly question the existence I assume is related to the axiom of choice, is that right?
Assume now that $S$ is a topological space. We define a topology on $S / \sim$ by declaring a set $U$ in $S / \sim$ to be open if and only if $\pi^{-1}(U)$ is open in $S$. Clearly, both the empty set $\emptyset$ and the entire quotient set $S / \sim$ are open. Further since $$ \pi^{-1} \left( \bigcup_{\alpha} U_\alpha \right) = \bigcup_{\alpha} \pi^{-1} \left( U_\alpha \right) $$ and $$ \pi^{-1} \left( \bigcap_{\alpha} U_\alpha \right) = \bigcap_{\alpha} \pi^{-1} \left( U_\alpha \right) $$
Second silly question why rigorously we have the two equalities above?
All the rest is easy (to me).
Update : Actually I can give an attempt... I'll prove the first equality, the second seems almost the same.
$$ x \in \pi^{-1} \left( \bigcup_\alpha U_\alpha \right) \iff \pi(x) \in \bigcup_\alpha U_\alpha \iff \exists \alpha \;\;\pi(x) \in U_\alpha \iff \exists \alpha \;\; x \in \pi^{-1} \left( U_\alpha \right) \iff x \in \bigcup_{\alpha} \pi^{-1} \left( U_{\alpha} \right) $$
The second is similar but instead of $\exists$ I'd use $\forall$.
Is it right?
Sending a point to its class is unrelated to AC: it's just $\pi(x)=\{y \in S: y \sim x\}$, which needs no choice at all. The set of classes $S{/}{\sim}$ is always well-defined. A function going back, picking one representative for each class does require AC in general, but that's not what we're talking about.
Yes, the equalities for inverse images (of any function BTW; these are not specific to out $\pi$ at hand) are well-known and have easy proofs. The one you gave for unions is fine and intersections are similar. Normally this would be covered in a set theory chapter of the text boook, or in earlier courses.