The usual proofs of decidability for monadic first-order logic proceed by showing that any sentence $\phi$ of this logic has the small model property, i.e. if $\phi$ is satisfiable, then it is satisfiable in a finite model (with bounded cardinality). My question is: can this be extended to sets of sentences, i.e. let $\Gamma$ be a set of sentences of monadic first-order logic. Does it follow that, if $\Gamma$ has a model, then it has a finite model? Does it change the result if we allow equality?
My hunch is that this is not the case, i.e. that it's not possible to extend this property to sets of sentences, but I couldn't think of any counter-example.
$$\{\exists x_1,..., x_n(\bigwedge_{1\le i<j\le n}x_i\not=x_j): n\in\mathbb{N}\}.$$
EDIT: Without equality: $$\{\exists x(U_1(x)\wedge...\wedge U_n(x)\wedge\neg U_{n+1}(x)): n\in\mathbb{N}\}.$$
EDIT THE SECOND: If you remove equality and disallow infinite languages, however, then the answer becomes yes - this is simply because if $\Sigma=\{U_1,..., U_n\}$ is a finite unary relational language and $\mathcal{M}$ is a $\Sigma$-structure, the monadic first-order theory of $\mathcal{M}$ is determined entirely by the set of subsets of $\{1,...,n\}$ which are realized by elements of $\mathcal{M}$.
Specifically, any structure in a language consisting of $n$ many unary predicates is equality-free-monadic-first-order-equivalent to one of cardinality at most $2^n$.