Does a finite first-order theory which has a model always have a finite model? I'm curious whether this is true or not:

Let T be a finite, first-order theory. If T has a model, then T has a finite model.

I would assume the answer is 'yes', but I wanted to make sure I haven't missed something obvious. The reason I believe this to be the case is that according to my understanding of the Lowenheim-Skolem theorem, the only way a countable first-order theory can 'force' a size onto its potential models is by including a countable number of constants which the model must have representations for. Therefore, if the theory is finite, then the number of constants it can 'force' to exist must also be finite, thereby ensuring the existence of a finite model (assuming any model can exist at all). Is anything wrong with my reasoning (apart from its relative informality)?

Let the language contain a single binary predicate symbol, to be interpreted as "less than."
Consider the theory $T$ with the following axioms:
1) There are at least two objects.
2) The relation $L$ is a strict total order.
3) For any $x$ and $y$ such that $L(x,y)$, there exists a $t$ such that $L(x,t)$ and $L(t,y)$.
Then all models of $T$ are infinite. This is the theory of dense linear order. It has nice model-theoretic properties. Cantor proved that any countable densely ordered set with no first or last element is order-isomorphic to the rationals under the usual order.  Similarly, a countable densely ordered set with first element and no last is order isomorphic to the rationals in $[0,1)$, with similar results for the other two possibilities.
