If the function is not surjective, then at least one element of the codomain has no pre-image. However, because F is a function, every element in the domain is mapped to something in the codomain. But since we know that F is injective, then we can't use an element in the codomain more than once in our mapping. That must mean that there is an infinite set of elements in the codomain if we can "skip" one and still be injective. I'm having trouble, however, translating my thoughts into a proof format.
The Axiom of Choice is not available to me for the purposes of this proof. Infinite for this question would be that it as has a countably infinite subset.
Thank you to all who addressed my question, both for your help and patience.