This is the literal question: let $L$ be the language that only contains the two place relation symbol $<$. Prove there is no $L$-theory that has precisely the well ordered sets as models.
I assume the $L$-structure must be well ordered with respect to the relation $a^M \leq b^M \iff (a^M <^M b^M) \vee (a^M = b^M)$ where the superscript $M$ denotes it's the interpretation in the $L$-structure $M$.
Here's my work so far:
Suppose such a $L$-theory $T$ does exist. Add the constants $c_1, c_2, ...$ to $L$ giving a new language $L'$. Define the $L'$-theory $T'$ by adding $B := \{c_{n + 1} < c_{n} \ | \ n \in \mathbb{N}\}$ to $T$.
Now $T'$ cannot be consistent, because then there exists an $L'$-structure $M$ such that $M \models T'$, so $M \models T$, so $M$ is a well ordered set, but also $c_1^M > c_2^M > ...$, which is a contradiction. By the contraposition of the compactness theorem, there must exist a finite subset $A$ of $T'$ that is inconsistent as well. This $A$ must contain a finite number of elements from $B$. I'm stuck here.
I thought it would be better to give an answer rather than carry on with the comments. Your argument contains the key ideas but doesn't quite fit together. To fix it, argue like this:
Let $T$ be any $L$-theory that has every well-founded set as a model. As you suggest, we can extend $L'$ by adding constants $c_n$ and consider the set of sentences $B = \{ c_n > c_{n+1} \mid n \in \Bbb{N} \}$. Then any finite subset of $T \cup B$ is consistent, since it has a model in the well-founded ordered set $\Bbb{N}$. By the compactness theorem, $T \cup B$ has a model. $M$ say. Since every sentence in $B$ holds in $M$, the $c_n$ form an infinite descending chain in $M$, so $M$ is not well-ordered. Thus any $L$-theory that has every well-founded set as a model also has models that are not well-founded. Hence there is no $L$-theory whose models are precisely the well-founded sets.