Let $\mathbb{N}$ be the set of all finite ordinals defined as the intersection of all ordinals including the empty set and closed under successor. Consider the following set:
$S_0 = \mathbb{N}$
$S_{i+1} = S_i - i$
$S_0 = \{0, 1, 2, 3 , ...\}$
$S_1 = \{1, 2, 3, 4, ... \}$
$S_2 = \{2, 3, 4, 5, ... \}$
...
It is easy to show the empty set is not a member of $S$. Assume $S_z$ is empty. Then $S_{z-1} = \{z-1\}$ and $z-1$ is the largest finite ordinal. It is also easy to see $S_{i+1} \subset S_i$. Is $S$ ill-founded? Can $S$ exist in a model of ZFC?
You seem to confuse $\in$ with $\subseteq$. The fact that the ordered set $(S,\subseteq)$ is not well-founded doesn't mean that $(S,\in)$ is not well-founded.
After all, if no well-founded orders could have been modeled using sets, how could we have modeled the real numbers using sets, or the rational numbers for that matter, or the integers for that matter?
You might also note that $S_0=\Bbb N$, but for all $i>0$ it is true that $S_i$ is not a finite ordinal. Therefore $S_i\notin\Bbb N$. Therefore $\Bbb N\cap S=\varnothing$. And since this is a decreasing chain of sets, it follows that for all $i$, $S_i\cap S=\varnothing$. And so in fact, you can prove that $(S,\in)$ is a discrete order, namely no point is an element of any other point.
The key point, again, is the fact that $\subseteq$ was never required to be well-founded. And yes, $S$ is easily definable in $\sf ZFC$, from $\Bbb N$ (which itself is definable without parameters):
$$S=\{A\subseteq\Bbb N\mid\exists n\in\Bbb N\forall k\in\Bbb N(k\in A\leftrightarrow n\leq k)\}$$
(Remember that for ordinals, $n\leq k$ if and only if $n\in k\lor n=k$ or alternatively, $n\subseteq k$, both are fully expressible in the language of set theory.)