From an axiomatic set theoric approach why can we take uncountable unions?
From an axiomatic set theoric approach why can we take uncountable unions in the first place? Sure if $A$ and $B$ are sets then the thing denoted by $A\cup B$ exists and is a set, but if we want to extend this to an uncountable union shouldn't we assume some form of transfinite induction (and hence $\sf AC$)?
The reason is simple. Unions are not defined "in sequence".
The axiom of union tells us that if we have a set $X$ (whose elements are sets, but that's a given in set theory), then $\{a\mid\exists x\in X: a\in x\}$ is a set, and we denote that set as $\bigcup X$, since it is exactly the union of all the members of $X$.
In the case where $X=\{A,B\}$ we normally write $A\cup B$ and not $\bigcup X$. But that is the abuse of notation, rather than the generalized union.
Remarks.
You will find absolutely no trace of choice here.
Transfinite recursion/induction does not require choice. It is just that most common uses of choice require transfinite recursion/induction.
Of course one might ask about intersection next. The same answer holds, in principle, although you do not need the axiom of union this time, you need the axiom of separation instead. And we need to require that the family is non-empty, since $\bigcap\varnothing$ is the entire set theoretic universe (consider that division by zero kind of error).
You can find additional support of this idea in the fact that many of the properties which hold for finite unions, and are usually proved by induction in introduction to discrete mathematics classes, are actually much simpler to prove directly for $\bigcup X$-style definitions. For example, DeMorgan laws (with the caveat that $X$ is non-empty).
No. All we do is assume the
Axiom of Union. If $A$ is a set then there exists a set $U$ such that $$ \forall x\colon (x\in U\leftrightarrow \exists y\in A\colon x\in y).$$