Prove that $|\cos(\sin(x_1)) - \cos(\sin(x_2))| \leq |x_1 - x_2|, \forall x_1, x_2 \in \mathbb R$. I asked this question without any limitation on methods that might be used. I believe it's turned out to be interesting to see a variety of different approaches. It turns out that the aim of the exercise that my students were given was to provide an example of the application of the Mean Value Theorem. I really like the MVT solution as it's a nice little application showing how powerful elementary calculus can be. None the less, I've enjoyed following other lines of thought.
I'm ticking Christian Blatter's solution as it's a particularly nice application of precalculus knowledge. It would be a good challenge question for brighter maths students in the final year of high school after introducing the various trigonometric formulae. It's a neat application of converting a difference to a product.
Finally, it's a good example of the contrast between pre and post calculus. Thanks to everyone who has contributed to this page.
 A: It suffices to prove that $|\sin b - \sin a| \le |b - a|$, because then we will have:
\begin{align*}
|\cos \sin x_1 - \cos \sin x_2|
&= |\sin (90 - \sin x_1) - \sin(90 - \sin x_2)| \\
&\le |(90 - \sin x_1) - (90 - \sin x_2)| \\
&= |\sin x_1 - \sin x_2| \\
&\le |x_1 - x_2|
\end{align*}
To prove $|\sin b - \sin a| \le |b - a|$ (i.e. $\sin$ is Lipschitz with constant 1), either refer to Christian Blatter's explanation, or else assume WLOG $a \le b$ and use calculus:
$$
|\sin b - \sin a| = \left| \int_a^b \cos x \; dx \right| \le \int_a^b |\cos x| \; dx \le \int_a^b 1 \; dx = |b - a|
$$
A: One has
$$|\sin x|\leq|x|\qquad(x\in{\mathbb R})$$
and consequently
$$|\cos\alpha-\cos\beta|=2\left|\sin{\alpha+\beta\over2}\right|\ \left|\sin{\alpha-\beta\over2}\right|\leq |\alpha-\beta|\ .$$
Similarly one proves $|\sin\alpha-\sin\beta|\leq|\alpha-\beta|$. Putting it together we obtain
$$|\cos(\sin x_1)-\cos(\sin x_2)|\leq|\sin x_1-\sin x_2|\leq|x_1-x_2|\ .$$
A: Set 
$y(x) = \cos (\sin x); \tag{1}$
then
$y'(x) = -(\sin (\sin x)) \cos x. \tag{2}$
Note that, since $\vert \sin x \vert \le 1$, we have
$\vert \sin (\sin x) \vert <1; \tag{3}$
in fact, there exists a positive real $k$ with $\sin 1 < k < 1$ such that
$\vert \sin (\sin x) \vert < k; \tag{3}$
(3) follows from the facts that $1 < \frac{\pi}{2}$ and $\sin$ is monotonically increasing on $[0, \frac{\pi}{2}]$.  From (3) we have, since $\cos x \le 1$ for all $x$,
$\vert y'(x) \vert = \vert (\sin (\sin x)) \cos x \vert = \vert (\sin (\sin x)) \vert \vert \cos x \vert < k \le 1, \tag{4}$
and now we simply integrate:
$\vert y(x_2) - y(x_1) \vert = \vert \int_{x_1}^{x_2} y'(s) ds \vert \le \int_{x_1}^{x_2} \vert y'(s) \vert ds   \le k \vert x_2 - x_1 \vert < \vert x_2 -x_1 \vert, \tag{5}$
provided $x_1 \ne x_2$, so substituting 
(1) into (5) yields
$\vert \cos (\sin x_2) - \cos (\sin x_1) \vert < \vert x_2 - x_1 \vert; \tag{6}$
the inequality is apparently strict unless $x_2 = x_1$, in which case the two sides of (6) take the same value $0$.  This proves the required inequality in all cases.  QED.
Wow!  Easier than I first thought it would be!
Hope this helps.  Cheers, 
and as always,
Fiat Lux!!!
A: Okay, the deadline has passed and the required solution for my students uses the Mean Value Theorem. I'll present that here, but there's more to it and I've edited the question as well to explain.
Calculus Solution Specifically Using the Mean Value Theorem
Let $f(x) = \cos(\sin(x))$ then $f'(x) = -\sin(\sin(x))\cos(x)$ which is defined for all $x \in \mathbb R$. So it is certainly continuous on any closed interval and differentiable on any open interval, so it satisfies the criteria for the Mean Value Theorem on $[x_1, x_2]$. So, there exists a $c \in [x_1, x_2]$ with
$$\frac{\cos(\sin(x_2)) - \cos(\sin(x_1))}{x_2 - x_1} = f'(c).$$
But, we can also see that $\left|f'(c)\right| \leq 1$ as $\left|\cos(c)\right| \leq 1, \forall\, c \in \mathbb R$ and $\left|\sin(\sin (c))\right| \leq 1, \forall\, c \in \mathbb R$, so together these imply
\begin{align*}
&\left|\frac{\cos(\sin(x_2)) - \cos(\sin(x_1))}{x_2 - x_1}\right| = \left|f'(c)\right| \leq 1 \\
&\Rightarrow \left|\cos(\sin(x_2)) - \cos(\sin(x_1))\right| \leq \left|x_2 - x_1\right|\quad\blacksquare
\end{align*}
