Of course that we need to assume that $\sf ZFC$ is consistent for these arguments, since if there are no models of $\sf ZFC$, then all the models are well-founded.
The usual answer is given by compactness, as bof writes in a very good answer. Here is a different argument, which relies on absoluteness and incompleteness (or well-foundedness, the last remark) instead.
The idea is that if $(M,\in)$ is a well-founded (and without loss of generality, transitive) model of $\sf ZFC$, then $V$ and $M$ agree on arithmetic statements, in particular they agree on what are the rules of first-order logic, the axioms $\sf ZFC$ have, and whether or not these axioms are consistent. Since $V$ is aware of $M$ being a model of $\sf ZFC$, it means that every well-founded model of $\sf ZFC$ must also agree that $\sf ZFC$ is consistent and has a model.
But by the incompleteness theorem we know that $\sf ZFC$ cannot possibly prove that it is consistent, so there must be a model $(N,E)$ which thinks that $\sf ZFC$ is inconsistent, and by the above it cannot be well-founded.
Caveats.
When we talk about $\sf ZFC$ being consistent as an arithmetic statement, we mean that there is a recursively definable predicate on the natural numbers which encodes the axioms of $\sf ZFC$. So when I say that $V$ and $M$ agree on what is first-order logic and what are the axioms of $\sf ZFC$, I mean that they interpret that predicate in the same way.
Of course, the model which thinks that $\sf ZFC$ is inconsistent will have non-standard integers, which means that he will think that $\sf ZFC$ is a larger theory and includes more axioms than it really has and the logic will have more inference rules and longer proofs, which is where the contradiction comes from.
The absoluteness argument shows that in fact if $N$ is non-well founded but still agrees on what $\omega$ might be (this is called an $\omega$-model), then $N$ still agrees on what is $\sf ZFC$ and whether or not it is consistent.
So the existence of $\omega$-models is itself stronger than just the existence of models of $\sf ZFC$. But it is weaker than the existence of transitive models, since those can be non-well founded.
You can get the model $N$ using well-foundedness instead. Order the transitive models of $\sf ZFC$ by $\in$, since it's a non-empty class (otherwise all models are non-well founded; or there are no models at all) it has a minimal element, $M$. But now $M$ agrees that $\sf ZFC$ is consistent, so there is some $(N,E)\in M$ such that $M\models(N,E)\text{ is a model of }\sf ZFC$. Now we know the fact that the satisfaction relation is absolute between transitive models, $(N,E)$ is a model of $\sf ZFC$. But since $M$ was a minimal transitive model, $N$ cannot be transitive.