Is this true? For any hyperfinite $n$ that isn't finite, there is a hyperfinite set $A$ such that $\mathbb R \subset A$ and $|A|\le n$ (that's the crucial part, of course)? Intuitively it seems right, but I haven't found a reference and I am not very good at NSA. 



Hyperfinite set containing the reals, with specified upper bound on internal cardinality?
This depends on how strong your axioms for non-standard analysis are. Certainly any sufficiently saturated model will have an $A$ of the sort you ask about. [Proof: The collection of formulas consisting of "$x$ is a set", "$|x|\leq n$", and "$r\in x$" for all standard reals $r$ is finitely satisfiable. So by saturation there is an $A$ satisfying the whole collection.) But there are various weaker assumptions than $(2^{\aleph\_0})^+$-saturation that are sometimes used in NSA and that might not suffice to provide the $A$ that you want.


