Injective continuous real-valued functions on an interval must be monotone

It is a well-known and very useful theorem of real analysis that injective continuous real-valued functions on an interval must be monotone. This is usually proved by contradiction using the Intermediate Value Theorem (IVT). However the direct negation of a function being monotone is actually not very friendly.

Let $I$ be a (nondegenerate) interval in $\R$.

A function $f:I\to\R$ is monotone increasing if (and only if), for all $x_1,x_2 \in I$ with $x_1\leq x_2$, we have $f(x_1) \leq f(x_2)$. Negating this we see that $f$ is not monotone increasing if there exist $x_1,x_2 \in I$ with  $x_1\leq x_2$ such that $f(x_1) > f(x_2)$. 

[Of course in this case we must actually have $x_1<x_2$.]

Similarly, $f$ is not monotone decreasing if and only if there exist $x_3,x_4 \in I$ with  $x_3\leq x_4$ such that $f(x_3) < f(x_4)$. 

[Again, in this case we must have $x_3<x_4$.]

Now $f$ is monotone (on $I$) if and only if $f$ is monotone increasing or $f$ is monotone decreasing. So we can now say what "not monotone (on $I$)" means:

A function $f:I\to\R$ is not monotone (on $I$) if and only if $f$ is not monotone increasing and $f$ is not monotone decreasing, and this happens if and only if there are two pairs of points in $I$, $x_1<x_2$ and $x_3<x_4$, such that $f(x_1) > f(x_2)$ and $f(x_3) < f(x_4)$. 

Note, however, that we have no idea here how these points are arranged, and in particular we do not know whether the intervals $[x_1,x_2]$ and $[x_3,x_4]$ overlap or not. Sorting out the cases just gets too ugly, I think, though every case you look at does lead to an easy contradiction using the IVT.

This isn't a very pleasant position to arrive at, so I do not recommend starting a proof by contradiction by saying (about our injective continuous function $f:I\to\R$) "Assume, towards a contradiction, that $f$ is not monotone on $I$", unless you have already planned an efficient follow-up.

What is your favourite way to get around this? A lot of the time I see proofs that claim that if a (continuous and injective) function $f:I\to\R$ is not monotone, then there must either be three points $x_1<x_2<x_3$  in $I$ with $f(x_1)<f(x_2)$ and $f(x_2)>f(x_3)$ (a sort of hat), leading easily to a contradiction of injectivity using the IVT, or else there must be three points $x_1<x_2<x_3$ in $I$ with $f(x_1)>f(x_2)$ and $f(x_2)<f(x_3)$ (a sort of inverted hat or a "check"), again leading easily to a contradiction of injectivity using the IVT. But although it is true that a non-monotone function must have three points like that, it isn't actually an immediate consequence of negating the definition of monotone.

One method (probably not the most sensible) that I thought up reminds me a bit of some applications I have seen of Ramsey Theory.

I consider the notion of $f$ being "monotone on a subset $S$ of $\R$", where $S$ need not be an interval. (The definition is still obvious.) Now, if a function $f:I\to\R$ has no hats or checks (in the sense above), that means that $f$ is monotone on every 3-subset $S$ of $I$ (i.e. every subset $S$ of $I$ such that $S$ has exactly 3 elements). It is then relatively easy to deduce that $f$ is monotone on every 4-subset of $I$. (It is even easier if you assume injectivity.) And now you can deduce that $f$ is monotone, because no points $x_1,x_2, x_3, x_4$ can exist as above. (Here the set $S=\{x_1,x_2,x_3,x_4\}$ must either be a 4-set or a 3-set. It can't be a 2-set because the strict inequalities go in opposite directions!)

Another, probably more sensible, approach is to first prove that $f$ is monotone on every closed subinterval of $I$. (The number of cases is dramatically reduced if you already have a particular inequality between the values at the endpoints.)

So, what is your favourite proof?