I know that, from compactness theorem, one can prove that there are models of first order arithmetic in which there is some "number" which is not a successor of zero, in the sense that it is strictly bigger than any successor of zero (i.e. any element of the model obtained by applying the successor function to zero finitely many times).

From the axiom of infinity, it follows that every model of ZFC must contain an element which one can think of as "the natural numbers", in the sense that it is a model of Peano Axioms. Peano Axioms are a second-order theory, since the principle of induction is a second order axiom, and from the principle of induction it follows that the Peano Axioms have a unique model in ZFC, so that we can call this model among the others of first-order arithmetic, the "standard" model of arithmetic.

But picking one model of ZFC, how do we know what's really inside its standard model of arithmetic? How do we know there is nothing else than the successors of zero? After all, every model of ZFC thinks that his natural numbers are the standard ones, so one can use compactness to produce a model of ZFC which has non-standard natural numbers from an external point of view, and in which there will be a standard natural numbers object containing elements bigger than any successor of zero.

So, if one wants to do mathematics inside a model of a first order theory of sets, how can one know that he is able to pick a model in which the natural numbers are not non-standard?

whatever axioms you likethat machine M halts or doesn't halt on input I. If your chosen axioms are consistent, then some choices of (M, I) must fail to have proofs, or else it would violate the halting problem. Then "Machine M, with input I, halts in $\alpha$ steps, and $\alpha \in \mathbb{N}$" is independent of your chosen axioms. $\alpha$ must be nonstandard, of course, because otherwise a proof would necessarily exist. $\endgroup$ – Kevin May 19 '19 at 19:44