The first-order properties of the reals are what we call "real closed field" ... so start with the reals, add an infinite element to that ordered field [say use the rational functions $\mathbb R(t)$ over $\mathbb R$ where $t$ is large and positive]. Then take its real-closure. So it is enough to show that the real-closure of an ordered field of power $\mathfrak c$ still has power $\mathfrak c$. [Concretely, perhaps, a space of Puisieux series.] Maybe this approach has a less "Axiom-Of-Choice" feel than either ultrafilter or Lowenheim-Skolem.
added Dec 14
I am not convinced by André's comment. But anyway, my answer is for the first-order theory of the ordered field of the reals with only these usual relation symbols: $+, \cdot, 0, 1, \lt$. Only the OP can tell us if this is what he really intended.