Let $\{A_i\mid i\in n\}$ be a finite family of infinite sets. ( That is, $A_i$ is infinite for every $i\in n$ and $n\in \mathbb{N}$)
Here, we can choose representative $a_i$ from each $A_i$ and construct $\{a_i\mid i\in n\}$.
This process really doesn't use Axiom of Choice?
How do I write this process down formally (in mathematical language)
$\newcommand{\Zobr}[3]{#1 \colon #2 \to #3}$ Claim: Let $\{A_i; i\in n\}$ be a system of non-empty sets, $n\in\omega$. Then there is a choice function $\Zobr fn{\bigcup A_i}$.
Proof. Induction on $n$.
$1^\circ$ For $n=0$ put $f=\emptyset$.
For $n=1$: Since $A_0$ is non-empty, we know that $(\exists a\in A_0)$. For every such $a$ the function $f=\{(0,a)\}$ has required property.
We have used existential instantiation here.
$2^\circ$ Suppose that there is a choice function $\Zobr fn{\bigcup A_i}$. We want to extend it to a new function $g$ defined on $n+1=n\cup\{n\}$ in a such way that $g(n)\in A_n$. Again we know that there exists $a\in A_n$ and we put $g=f\cup\{(n,a)\}$. (Again, this is application of existential instantiation.)
