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.

This depends on how strong your axioms for nonstandard 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. 


Andreas has pointed out that in any sufficiently saturated nonstandard model $\mathbb{R}^\ast$, the answer is yes. Meanwhile, let me point out that if, as is commonly done, one builds one's hyperreals $\mathbb{R}^\ast$as the ultrapower of $\mathbb{R}$ by an ultrafilter on $\mathbb{N}$, then the answer is no. Indeed, in such a nonstandard $\mathbb{R}^\ast$, there is no hyperfinite cover of the standard reals at all. To see this, suppose that $x$ is a nonstandard hyperfinite set of reals. In the ultrapower, $x$ is represented by a function $f$ from $\mathbb{N}$ to the finite sets of reals, so that $f(k)$ is a finite set of reals. By the Los theorem, the reals $r$ with $r^\ast\in x$ must all have $r\in f(k)$ for almost all $k$. But this is a countable set, so $x$ contains at most countably many reals. 

