Can I use the fact that $\mathbb{N}$ is a total order to prove that it is infinite? We can show the canonical ordering $\le$ gives a total ordering on $\mathbb{N}$.  We can show that any finite subset of a totally-ordered set has a maximal/greatest element.  For brevity, take these as given.  Assuming $\mathbb{N}$ finite, we then have a maximal element $M \in \mathbb{N}$.  But by successor axiom of $\mathbb{N}$, we also have $M+1 \in \mathbb{N}$.  This implies $M+1 \le M$, which contradicts our ordering.
Is there anything wrong with this approach?  I have seen other proofs which seem slightly more tedious, so I was wondering perhaps there is something I might be missing?

In my opinion, there is something wrong. Obviously, your proof works, however when you use the total order, do not forget that it has been constructed using the successor axiom, furthermore you use this axiom in your proof. That is, on the whole, the infiniteness is rather a consequence of the successor axiom than the total order. 
To state the obvious, finite sets have a total order so total order does not imply something on the cardinality.
It depends on the precise axioms you used to construct $\mathbb{N}$, but I think that you can show (by induction) that no subset $\{1,...,n\}$ is in bijection with $\mathbb{N}$ in a more elementary way.
