Is induction strictly required to prove finite AC? Consider the following proof of the (finite) AC in ZF. Assume we are given a finite, non-empty collection $\mathcal{F}$ of non-empty sets. Hence $\mathcal{F}= \{A_i : 1 \leq i \leq n\}$ for some positive integer $n$. Since the sets are non-empty, we have that $\exists x_1 \exists x_2 \cdots \exists x_n \ \forall i \in \{1,2,\cdots ,n\} \  x_i \in A_i$. Hence by existential instantiation $y_i \in A_i$ (for some new symbol $y_i$). Define the choice function $A_i \mapsto y_i$. I have seen variants of this proof for instance on MathOverflow.
Some time ago I recall seeing a comment here by Asaf Karagila (which I cannot cite off the top of my head) which states that the above argument is in fact incorrect for some technical reason related to non-standard integers and some issue related to Skolem's paradox, and that the "correct" proof of finite AC in ZF requires induction.
I would ask for a detailed explanation of the problem with the above proof, which I still don't quite understand. I would also like to know how induction "saves the day" so to speak: isn't the proof in the first paragraph just the inductive proof "unrolled"?
 A: To prove that finite choice holds in $M$, some model of $\sf ZF$, you need to show that if $$M\models A\text{ is a finite family of non-empty sets},$$ then $$M\models A\text{ admits a choice function}.$$
Your argument is to use induction in the meta-theory and simply repeatedly apply Existential Instantiation to pick one element at a time from one set in $A$ at a time.
But if $A$ happens to have non-standard size, assuming we have non-standard integers, then it is in fact larger than any possible length of a proof, since proofs live in the meta-theory, so they only have standard lengths.
So how will your induction help you to show that $A$ has a choice function? Instead, we can utilise internal induction, and the proof goes through just fine since we only use Existential Instantiation twice (or at least a very small number of times) in the induction step. Then relying on the fact that the internal induction gives us that $M$ satisfies finite choice.
Of course, we can formalise the whole thing inside a non-standard meta-theory, but we can ignore that for all intents and purposes, since we were not planning to write down the actual proof. Indeed, the idea of iterated Existential Instantiation is merely showing that we can find longer and longer proofs for larger and larger fragments, but "finite choice" is one statement, and should have one proof. So this should have been a hint from the get go.
