I'm not sure how to phrase the question, however I've been pondering on this idea for a while.
Perhaps the example will clarify what I mean:
The real number $\mathbb R$ can be defined axiomatically, they're the complete arquimedian ordered field (I may have forgotten some property).
In order to prove that there exists a model of the real numbers, I constructed an ordered set with two operations +, • and proved that if there exists a model of the real numbers, then I constructed a structure preserving bijection (preserves order addition and multiplication) $\Phi: \mathbb R \to R$, from which we can conclude that $R$ is an ordered field, with identities $0 = \Phi(0)$ and $1 = \Phi(1)$, and that $\Phi$ is an ordered field isomorphism, and thus R is a construction of $\mathbb R$.
However, this all hinged on the fact that $\mathbb R$ has a model, but even so R should still be an ordered field that satisfies all the axioms of the reals, is it therefore valid to say that $\mathbb R$ has a model and that R is one of them, by proving it is if a model exists?
I'm sorry for the poor formatting, I'm writing this on my phone. If this is false can anyone give a counterexample?