Let's consider the signature of the ordered sets theory as having just two binary relations $=$ and $\leq$ and no functions. Now, consider whether we can express the well-foundedness property as a formula of first-order logic.
What I've come up with is $$ \lnot \exists x(\exists y(y < x) \land \forall y ((y < x) \rightarrow \exists z (z < y))), $$ where I also took the liberty to use $x<y$ as a shorthand for $(x \leq y) \land \lnot (x = y)$.
Essentially, this is a negation of the statement that there exists an infinitely decreasing sequence. But I've been hinted that this might not be expressible in first-order logic, so am I right, or did I make a mistake?
Take the integers $\mathbb{Z}$ ordered in the usual way and to each $n \in \mathbb{Z}$ adjoin an element $n'$ such that $n' < n$. More formally, define a partial order on $\mathbb{Z} \times \{0, 1\}$ by $$(n, s) < (m, t) \iff (s = t = 1 \land n < m) \lor (n = m \land s = 0 \land t = 1).$$
This relation satisfies your formula but is not well-founded.