Non-standard model of $Th(\mathbb{R})$ with the same cardinality of $\mathbb{R}$ Let $\mathfrak{R}= ⟨\mathbb{R},<,+,-,\cdot,0,1⟩$ be the standard model of $Th(\mathbb{R})$ in the language of ordered fields. I need to show that there exists a (non standard) model of $Th(\mathbb{R})$ with the same cardinality of $\mathbb{R}$ but not isomorphic to $\mathfrak{R}$.
My doubts are on how to prove the "not-isomorphic" part, so I'll just sketch my proof of existence of a model: 
Let's expand the language adding a new constant $c_x$ for each $x \in \mathbb{R}$ and an additional new constant $c$. We consider the following set of formulas:
$$ \Gamma = Th(\mathbb{R}) \cup \{c_x \ne c_y\}_{x,y \in \mathbb{R}, x \ne y} \cup \{c_x < c\}_{x \in \mathbb{R}}$$
We can show $\Gamma$ is finitely satisfiable, thus, by Compactness Theorem, satisfiable, thus it has a model with cardinality $\kappa \ge |\mathbb{R}|$. By the Löwenheim-Skolem Downward theorem it has a model of cardinality $c = |\mathbb{R}|$, say $^*\mathfrak{R}$.
Now my doubts are how to show that $^*\mathfrak{R} \not \cong \mathfrak{R}$. I guess the lack of an isomorphism should be showed after shrinking back the structures to the original language, i.e. removing all the constant symbols. Moreover, as the two models have the same cardinality we cannot simply conclude there is not a bijection, so I think a possible proof is assume there is such an isomorphism and get a contradiction with the order relation, but all my reasonings in this direction seem at a dead point. How can one show there is not such an isomorphism?

Instead of compactness, you can try an ultrapower argument. 
Let $\cal U$ be a nonprincipal ultrafilter on $\Bbb N$. Then $\Bbb{R^N}/\cal U$ is elementarily equivalent to the reals. And it is not hard to show it has the wanted cardinality, and that it is non-standard. 
