So we were given this example to show that substructure that is elementary equivalent to its structure, isn't necessarily an elementary substructure.
Just so I'll be clear on the definition: A substructure $M$ of $N$ is an elementary substructure, if for every formula $\phi (v_1, \cdots, v_n)$ and for every $\overline b=(b_1, \cdots, b_n) \in M^n$, $M \vDash \phi( \overline b) \iff N \vDash \phi( \overline b)$.
So the example we were given is $N=(\mathbb N; <)$ and $M=(5,6,7,...;<)$. So there is an isomorphism $f(n)=n+5$, so $N$ and $M$ are elementary equivalent. However, and this is the part I don't get, for $\phi: \exists x: x\lt 5$, indeed $N \vDash \phi$ but $M \not\vDash \phi$. So $M$ isn't an elementary substructure of $N$.
Which is true in general, for $\overline b \in N^n$, but our demand to prove (or refute), is that $\overline b \in M^n$, and there is no such $\overline b$ so that $N \vDash \phi (\overline b)$, or am I missing something here?
The given formula $\phi:=\,\exists x: x<5$ has no unbounded variable, but it uses a constant (namely $5$) which is not available in the language, as we only have a binary operation symbol ($<$) and no constant symbols.
As nombre commented, the formula was most probably with a free variable: $$\phi(y):=\,\exists x:x<y\,.$$
Now we indeed have that $f=n\mapsto n+5$ is an isomorphism between the structures $N$ and $M$, which implies that $N$ and $M$ are elementary equivalent (they validate the same closed formulas on the given language).
However, choosing $b=5\in M\subseteq N$ with $\phi$ in the definition of 'elementary substructure', we will have $N\models\phi(5)$ while $M\not\models\phi(5)$.