An ordered field with a infinitesimal element elementarily equivalent to the reals Let $\mathcal{L} := \{0, 1, +, -, \cdot, \leq \}$ be a language, where $0, 1$ are constants and $+, -, \cdot$ are binary functions and $\leq$ is a binary relation. 
We consider the ordered fields in this language. An element $\varepsilon$ is said to be infinitesimal if $0 < \varepsilon < \frac{1}{n}$ for all integers $n \geq 1$. 
The exercise is now the following: Show the existence of an ordered field $K$ such that $K$ admits an infinitesimal element and is elementarily equivalent to $\mathbf{R}$. 
First I thought it might be good to mimic the construction of the hyperreals, but I do not know whether they are elementarily equivalent to $\mathbf{R}$ and the construction looks quite lengthy.
Can anyone give me a hint on how to start the proof? 
Thanks!
 A: After the very helpful discussion in the comments, I will try and put a whole answer here (if any of the commenters want to post an answer, please do so and I'll accept it): 
Let $\mathcal{L}' := \mathcal{L} \cup \{ \varepsilon \}$ with $\varepsilon$ a constant. Let $T' := Th(\mathbf{R}) \cup \{ 0 < \varepsilon < \frac{1}{n} \}$. 
We use the compactness theorem: A theory $T$ is consistent if and only if every finite subset of $T$ is consistent. 
Let $T_0$ be a finite subset of $T'$. There is a $N \geq 1$ such that for any axiom $0 < \varepsilon < \frac{1}{n} \in T_0$ we must have $n \leq N$. The reals form a model of $T_0$ (they clearly satisfy the axioms of $T$ and for $\varepsilon$ take $\frac{1}{N  +1}$). 
By the compactness theorem $T'$ is consistent. Thus there exists a model $K$ for $T'$. 
A fortiori $K \models T$ for $K$ viewed as an $\mathcal{L}$-structure and $Th(\mathbf{R}) \subset Th(K)$. Let $F \in Th(K)$. We have $\mathbf{R} \models F$ or $\mathbf{R} \models \neg F$. But if $\mathbf{R} \models \neg F$, then $\neg F \in Th(K)$. This is of course a contradiction, hence $Th(\mathbf{R}) = Th(K)$. 
Added: I found this math overflow link: https://mathoverflow.net/questions/39504/what-are-examples-of-ordered-fields-that-do-not-have-the-archimedean-property, which gives the explicit example $\mathbf{R}(X)$ for an ordered field with infinitesimal element $\frac{1}{X}$.
A: Ordered fields elementarily equivalent to the reals are called real closed fields.
Any ordered field $F$ has a real closure $F^r$ which is an algebraic extension that is real closed.
So, if you let $F$ be any non-archimedean ordered field, then $F^r$ is an example.
A standard, explicit construction of examples is that if $F$ is any real closed field, then so is the field of Puiseux series over $F$.
The field of Puiseux series over $F$ is the real closure of the field $F((t))$ of Laurent series over $F$, ordered so that $t$ is a positive infinitesimal.
