Why do we show that structures aren't isomorphic by exhibiting a property not shared by one of them? If someone asks me how to prove that two order structures $\langle A,\leq \rangle$ and $\langle B,\preceq \rangle$ are isomorphic I would immediately suggest: try to find a function $f:A\to B$ such that $a \leq b$ iff $f(a) \preceq f(b)$, for all $a,b \in A$. Because given the definitions, this is the natural way the proof goes.
But if the same person asks me to show that $\langle A,\leq \rangle$ and $\langle B,\preceq \rangle$ aren't isomorphic, then I'm in trouble. I would start suggesting him: (1) pick up an arbitrary function $f:A\to B$ such that $a \leq b$ iff $f(a) \preceq f(b)$, for all $a,b \in A$, (2) try to find a contradiction. Then most of the times this is where I get stuck. Because I don't find any further information than this generic assumption, and I simply cannot check all possible functions.
This started with my (failed) attempt to show that $\langle \mathbb{N},< \rangle$ and $\langle \mathbb{Z},< \rangle$ aren't isomorphic.
Then I googled for some answers and I realize most people prove such "$X$ and $Y$ aren't isomorphic" statements by showing a property of $X$ (or $Y$) that is not satisfied by $Y$ (or $X$). Like:

$\langle \mathbb{N},< \rangle$ is well-ordered, $\langle \mathbb{Z},< \rangle$ isn't. QED.

Question: Why is this a legitimate approach? Why is it enough? As a logician, and looking at the definitions, my natural approach to show that two structures aren't isomorphic, is to assume a generic order-preserving mapping and then deriving a contradiction. How is this "property approach" in harmony with that?
 A: I think contrapositive is nicer than contradiction. 
\begin{align*}[\text{isomorphism}] &\implies [\text{some property}],\text{ so}\\ [\text{lack of that property}] &\implies [\text{no isomorphisms}].\end{align*}
I personally had trouble separating the two: I would mistakenly think I had argued by contradiction, when in fact it was a contrapositive argument. Or, worse still, I would sully a nice contrapositive argument so it became a convoluted argument by contradiction.
So, rather than argue $$[\text{isomorphism}] \implies \text{contradiction},$$
it's mathematically 'cleaner' (less confusing, more fulfilling) in my opinion to use a property preserved by isomorphism in a contrapositive argument.

To elaborate, some proofs by contradiction are not ideal because they're unnecessarily bloated, as they already contain enough logic for a free-standing proof of the contrapositive. This is what I didn't realize for several years.
For example, examine a possible contradiction argument. You begin by assuming that the objects are isomorphic, hence at least one isomorphism exists and you pick one. This isomorphism must preserve certain properties. But you notice that a certain preserved-by-isomorphisms property holds for only one of the two objects! This clearly contradicts the existence of your chosen isomorphism, hence no isomorphisms exist, hence the two are not isomorphic.
That which was to be demonstrated has been beaten like a dead horse, because we observed that 


*

*Any isomorphism preserve certain properties

*One such property is not shared by both objects.
These two observations alone are enough for a contrapositive argument; it didn't need to be wrapped in a proof by contradiction.
A: This is because isomorphic structures are elementary equivalent (in fact, every isomorphism is an elementary embedding). Notice that the converse is not true. For example, there are many orders which are elementary equivalent, but not isomorphic, to $(\mathbb{N},<)$.
Another important method to prove the non-existence of an isomorphism is the usage of functors in the sense of category theory. It is a general fact that functors preserve isomorphic objects. This is especially useful in algebraic topology, where one uses several functors $\mathsf{Top} \to \mathsf{Ab}$ (and similar) in order to distinguish topological spaces.
A: The "property approach" does generate a contradiction. The full proof tends to follow these lines:
Suppose the structures $X$ and $Y$ are isomorphic. Then there exists an isomorphism $f : X \rightarrow Y$ between them. Property $P$ is preserved under isomorphism (due to some other proof), so because $X$ has property $P$, so must $Y$. But $Y$ does not have property $P$. Therefore, $X$ and $Y$ are not isomorphic.
