Here are some examples that might help in understanding. If ZFC is consistent, then it follows we have a set model $M$ of the theory. Consider a nonprincipal ultrafilter $U$ on $\omega$ and let $M^{\omega}/U$ be the induced ultrapower. $M^{\omega}/U$ is a model of ZFC, but it cannot be well-founded because its $\omega$ has nonstandard elements. Specifically, for any strictly increasing function $f: g: \mathbb{N} \rightarrow \mathbb{N}$ and $n \in \mathbb{N}$, we have $M^{\omega}/U \models \omega > (f)_{\mu} g)_{U} > n$ where the $\omega$ here is of course the nonstandard one.
Also, if you don't want to work with ultrapowers directly, you can simply appeal to the Compactness theorem. Introduce a set of constants $\{c_n| n \in \mathbb{N}\}$ into your language and for each $n \in \mathbb{N}$, define $\varphi_n := c_0 \ni c_1 \ni \ldots \ni c_n$. If ZFC is consistent, then every finite fragment of $ZFC + \{\varphi_n| n \in \mathbb{N}\}$ is consistent so the entire theory is by the Compactness theorem. This then gives us a model $M$ N$ of ZFC that externally can be seen to have an infinite descending chain:
$c_0 \quad \ni_M ni_N \quad c_1 \quad \ni_M ni_N \quad c_2 \quad \ni_M ni_N \quad \ldots \quad \ni_M ni_N \quad c_n \ldots$.
Note this is not an actual infinite $\in$-descending chain as Stefan points out. , but merely a binary relation on the set $N$. For example, if we have $x \notin M$ such that N = M^{\omega}/U$ is the ultrapower induced by a nonprincipal ultrafilter $\{x\} \in M$, we have U$ on $M$ thinking that \omega$, then $\{x\}$ is empty (i.e. \in_N$ would be defined by $\forall y(y (g)_U \quad\notin_M quad\in_N\quad(h)_U$ exactly when $\{n \quad\{x\}$)).in \mathbb{N}| g(n) \in h(n)\} \in U$.
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Clearing misconceptions: Defining “is a model of ZFC” in ZFC
