Dense substructure of countable dense linear order without endpoints Let $\mathbf A = (A,\leq)$ be a countable dense linear order without endpoints
and $\mathbf B \leq \mathbf A$ (a substructure of $\mathbf A$) which is dense in $\mathbf A$.
The problem: Show that $\mathbf B$ is an elementary substructure of $\mathbf A$.
My immediate tentative was to notice that if $\mathbf B$ is dense in $\mathbf A$, then $B$ is also countable, and so $\mathbf B\cong\mathbf A$.
Then I concluded that $\mathbf B\preceq\mathbf A$. 
However, I then realised this reasoning is wrong.
For example, $(2\mathbb Z,+,0)$ is a substructure of $(\mathbb Z,+,0)$, and they are isomorphic.
Nevertheless, the former is not an elementary substructure of the latter.
As an example of a sentence witnessing this, 
$$\exists x (x+x\approx2)$$ 
is valid on $\mathbb Z$ but not on $2\mathbb Z$. 
I suppose the trick must be in some cleaver use of the fact that $\mathbf B$ is dense in $\mathbf A$, although honestly, it's not even obvious to me that that is crucial (wouldn't, for example, $\mathbb Q\setminus[0,1]$ be an elementary substructure of $(\mathbb Q,\leq)$?).
I must also add that I only have the definition and no result to help me (in case someone says "Well, it's an immediate consequence of result X").
Thanks in advance!
 A: Well, it's an immediate consequence of quantifier elimination for $\text{DLO}$. If $B$ is dense in $A$, then also $B\models \text{DLO}$, so $B$ is an elementary substructure, since quantifier elimination implies model completeness.
Sorry, that was a bit facetious - you explicitly asked for an answer which is not of that form.
If you know the Tarski-Vaught test, you can apply that here. If you don't know it, you can reprove it by proving directly that for all $\varphi(\overline{x})$ and all $\overline{b}\in B$, $B\models \varphi(\overline{b})$ if and only if $A\models \varphi(\overline{b})$ by induction on the structure of $\varphi$. 
In either case, you'll end up having to check the case of a single existential quantifier (this is the hypothesis of Tarski-Vaught and the only tricky step in the induction proof). Here you'll need to use density of $B$, and the fact that if $c < a < c'$ and $c < b < c'$ in $A$, then there is an automorphism of $A$ fixing $(-\infty,c]$ and $[c',\infty)$ and moving $a$ to $b$ (this step is probably why the countability assumption on $A$ was included). 
Incidentally, you can use a very similar argument to prove quantifier elimination.
In actuality, the assumption that $A$ is countable is unnecessary, and as you suggested, the assumption that $B$ is dense in $A$ is unneccessary - all you need is that $B$ is dense in itself. So $\mathbb{Q}\setminus [0,1]$ is an elementary substructure of $\mathbb{Q}$, but $\mathbb{Q}\setminus (0,1)$ isn't (it's not a dense linear order). You can see these things directly from the model completeness argument.
