From Spivak's Calculus:
Prove that $|\sin x - \sin y| < |x - y|$ for all $x \neq y$. Hint: the same statement, with $<$ replaced by $\leq$, is a straightforward consequence of a well-known theorem.
Now, I might even be able to prove this somehow (?), but I can't seem to figure out what "well-known theorem" the author is alluding to here... any hints?
You don't actually need Calculus to prove it:
$$|\sin x - \sin y| = \left| 2 \sin \frac{x-y}{2} \cos\frac{x+y}{2} \right| \,.$$
The inequality $\left| \sin \frac{x-y}{2}\right|< \left|\frac{x-y}{2}\right|$ is well known, while $\left|\cos\frac{x+y}2\right|\leq 1$ is even more well known. The first inequality is sharp if $x-y \neq 0$.
If $x<y$ then one has $$\left|{\sin y-\sin x\over y-x}\right|=\left|{1\over y-x}\int_x^y\cos t\>dt\right|\leq\int_0^1 \bigl|\cos\bigl(x+\tau(y-x)\bigr)\bigr|\>d\tau<1\ ,$$ because the integrand is $\leq1$, but not $\equiv1$.
Maybe it's referring to the mean value theorem.
Define $f(x)=\sin x$.
Start with the case $y>x$ :
Note that $f(x)$ is everywhere continuous and differentiable, in particular in the interval [x,y] (y>x). By the mean value theorem (sometimes referred to as Lagrange's rule) $\exists$ a point $c\in(x,y)$ such that $$f(y)-f(x)=f'(c)\cdot(y-x)\ ,$$ i.e. $$\sin y-\sin x=\cos c\cdot (y-x)\qquad\Rightarrow\qquad |\sin y-\sin x|=|\cos c|\cdot | y-x| \ .$$ But $|\cos c|\leq 1$ hence $$|\sin y-\sin x|=|\cos c|\cdot | y-x|\leq |y-x|\ .$$
For the case $x>y$ the argument is analogous to the $y>x$ case. For $x=y$ it is trivially true.
