Consider the usual language and axioms of ZF. Now add constants $x\_1, x\_2, \dots$ to the language together with the axioms $x\_2\in x\_1, x\_3\in x\_2, \dots$ to form a new theory. Then by the compactness theorem, since every finite subset of the axioms has a model, the new theory has a model. But doesn't the set {$x\_1, x\_2, \dots$} have no $\in$-minimal element, contradicting the axiom of foundation?


I'm thinking that maybe {$x\_1, x\_2, \dots$} is not necessarily a set in the model, but isn't it by replacement? Maybe not, since we don't necessarily have a copy of $\mathbb{N}$ in our model... Could someone clarify this please?



Contradiction to axiom of foundation
You have already answered your question yourself: the set $\{x\_1, x\_2, \ldots\}$ cannot exist in the model, since this would violate the Foundation Axiom. You cannot get this set from Replacement, since there is no definable function having this set as its range, required to invoke the Replacement Axiom. 


What you have done is exactly to prove that if ZF is consistent, then there are non-wellfounded models of set theory. Your proof is one way to justify nonstandard analysis. 


Note that the ordinals of any model of your theory will have a non-wellfounded collection of ordinals, since the Levy rank of the sets x\_n will be decreasing.


