Is there an example of an ordered ring that is not isomorphic to any subring of the real numbers? Somebody told me it's possible to show the existence of rings non-isomorphic to any subring of the real numbers using model theory, but they couldn't get an example.
 A: An example of non-archimedean field via ultraproducts. The cartesian product of fields
$${\Bbb R}^{\Bbb N} = \Bbb R\times\Bbb R\times\Bbb R\times\cdots$$
isn't a field (isn't a model of...) because has zero divisors:
$$(0,1,0,1,\dots)(1,0,1,0,\dots)=(0,0,0,0,\dots).$$
The solution is taking a quotient: let be $\mathcal U$ a nonprincipal ultrafilter on $\Bbb N$. Define
$$(a_1,a_2,\dots)\sim(b_1,b_2,\dots)$$
when
$$\{n\in\Bbb N\,|\,a_n=b_n\}\in\mathcal U$$
The quotient ${}^*\Bbb R = {\Bbb R}^{\Bbb N}/{\sim}$ will be a field with infinitesimals, like the class of equivalence of the sequence
$$(1,1/2,1/3,1/4,\dots).$$
A: Another example (of a non-Archimedian ordered field) worth checking out is David Hilbert's example in Section 12 of Foundations of geometry. 
It is, to paraphrase, what you get by taking $\mathbb Q$, a single variable $t$, and then generating a field that is closed under the additional operation $x\mapsto \sqrt{1+x^2}$.
Viewed as real-valued functions, it can be proven that each such expression is eventually positive or eventually negative, and then the ordering is that $f\leq g$ if there exists an $b\in \mathbb R$ such that $f(x)\leq g(x)$ for all $x\in [b,\infty)$.
Hilbert's original description is a little archaic to a modern reader, but it's not too bad, and I'm sure there are more modern versions of it lying around somewhere.
A: Via model theory this is trivial: the theory of ordered rings is first-order, so in particular via Löwenheim–Skolem it has models of every infinite cardinality.
But this is much more elementary. Simply take your favourite ordered ring $R$, and consider the polynomial ring $R[X]$, and order it lexicographically${}^\dagger$. Then $R[X]$ is an ordered ring which is non-Archimedean: for every $n$ you have $n\lvert X\rvert<1$, but $X\neq 0$. This is not possible in a subring of real numbers (not with the standard ordering).
In fact, the same construction can be used to construct ordered rings of arbitrary (infinite) cardinality $\kappa$: you can just extend the order on $R$ to an order on the ring of polynomials in $\kappa$ variables, and for $\kappa> 2^{\aleph_0}$, this ring will not even be algebraically isomorphic to a subring of real numbers, simply because there are not enough reals.
In a way, this is an overkill: you can show that already ${\bf R}[X]$ is not algebraically isomorphic to a subring of the real numbers, even though it has the right cardinality. This is because ${\bf R}$ knows its own ordering: the positive elements are exactly the squares. In particular, the only ring homomorphism ${\bf R}\to {\bf R}$ must be constant, and thus onto (it must be constant on rationals for trivial reasons, and for any other real, the image is determined by the image of rationals below it). It follows that any ring homomorphism ${\bf R}[X]\to {\bf R}$ is not injective.

$\dagger$: As suggested by user49640 in the comments, when $R$ is a field, the ordering in $R[X]$ can be described another way: we can identify a polynomial $p\in R[X]$ with the polynomial function and say that $p<q$ if $p(r)<q(r)$ asymptotically as $r\to 0^+$.
Note also that we could have very well looked at asymptotic behaviour as $r\to +\infty$, and that would yield a different ordering of $R[X]$, which is also non-Archimedean, since then for any $n$ we would have $n\cdot 1<\lvert X\rvert$. This corresponds to the lexicographic ordering with the coefficients written starting from the higher orders, as suggested by KCd in the comments.
The two orderings of $R[X]$ can also be intuitively described as simply considering $X$ to be an infinitesimal or (respectively) infinitely large element. That way, it is clear that it is non-Archimedean.
