Is there an ordered field with distinct subfields isomorphic to the reals? Is there an ordered field with distinct subfields isomorphic to the field $\mathbb R$ of real numbers?
 A: Yes.  Let $K$ be any real-closed field that contains $\mathbb{R}$ as a proper subfield.  In particular, then, $K$ has a nonzero infinitesimal element $\epsilon$.  Let $B$ be a transcendence basis for $\mathbb{R}$ over $\mathbb{Q}$ and let $B'=\{b+\epsilon:b\in B\}$.  Then $B'$ is still algebraically independent (a polynomial with coefficients in $\mathbb{Q}$ evaluated at elements of $B'$ is infinitesimally close to the evaluation at the corresponding elements of $B$), and in fact there is an isomorphism of ordered fields $\mathbb{Q}(B)\to\mathbb{Q}(B')$ mapping $b$ to $b+\epsilon$ for each $b\in B$.
Now let $L$ be the algebraic closure of $\mathbb{Q}(B')$ in $K$.  Since $K$ is real-closed, this means $L$ is a real closure of $\mathbb{Q}(B')$ as an ordered field.  Also, $\mathbb{R}$ is a real closure of $\mathbb{Q}(B)$.  Since real closures are unique up to isomorphism and $\mathbb{Q}(B)\cong\mathbb{Q}(B')$ as ordered fields, this means that $L\cong\mathbb{R}$.  Thus $K$ contains two distinct subfields isomorphic to $\mathbb{R}$, namely $\mathbb{R}$ and $L$.
(In fact, $K$ contains $2^{2^{\aleph_0}}$ such subfields, since you could modify $B'$ to add $\epsilon$ only to some particular subset of $B$, and there are $2^{2^{\aleph_0}}$ different such subfields.  Or, you could get different such subfields by picking a different infinitesimal element to be $\epsilon$.  Since there is no bound on the number of infinitesimal elements such a field $K$ can have, there is no bound on the number of subfields isomorphic to $\mathbb{R}$ that an ordered field can have.)
