Let $\langle \mathbb{R}, 0, 1, +, \cdot, <\rangle$ be the standard model for $R$, and let $S$ be a countable model of $R$ (satisfying all true firstorder statements in $R$). Is it true that the set $1,1+1,1+1+1,\ldots$ is bounded in $S$? My intuition says "no", but I am yet to find a counter example. I read something about rational functions, but I cannot verify it is, indeed, a nonstandard model of R.

If $S$ is the set of real algebraic numbers then $1, 1+1, 1+1+1, \dots$ is unbounded in $S$. On the other hand, by compactness of first order logic (as Juris points out), there are models $S$ for which $1, 1+1, 1+1+1, \dots$ is bounded. 


Take for $S$ the field $F(t)$ with $F$ being the algebraic closure of $\mathbb Q$ inside $\mathbb R$. Equip it with the unique ordering for which $tx>0$ for every integer $x$, and take a maximal ordered algebraic extension of $F(t)$. The field $S$ that you get is realclosed, i.e., it has the same firstorder theory as $\mathbb R$. It is countable, and has nonarchimedean elements ($t$, for instance). 

