Show that Total Orders does not have the finite model property I am not sure whether my answer to this problem is correct. I would be grateful if anyone could correct my mistakes or help me to find the correct solutions.
The problem:  

Show that Total Orders does not have the finite model property by finding a sentence A which
  is refuted only in models with an infinite domain.

Just for reference purpose only:
A theory T has the finite model property if and only if whenever $T\nvdash A$ there is a model $\mathcal{M}$ with a finite domain, such that $\mathcal{M}$ satisfies the theory $T$ but $A$ does not hold in $\mathcal{M}$.
The theory of Total Orders (in the language with quantifiers, propositional
connectives, identity and one new binary relation symbol '<') defined as the set of consequences of the following three formulas:
1. $(\forall x)\neg(x<x)$
2. $(\forall x)((x<y\wedge y<z)\supset x<z)$
3. $(\forall x)(x<y\vee x=y\vee y<x)$
My answer is quite simple, but it is too simple that I doubt whether it is correct. I am trying to say that the statement "there is a least element" is refuted only in models with an infinite domain, for example the integers. My sentence A is $\neg((\exists y)(\forall x)(y<x))$.
I am not sure whether I am correct. Please correct me if I am wrong and please say so if there is any better answer.
Many thanks in advance!

Looks like a good candidate to me. As you say, this clearly holds in every finite model of our theory, but infinite counterexamples exist, like $\mathbb{Z}$.
