Answer
$\vert sin~a-sin~b \vert \leq \vert a-b \vert~~~~$ for all $a$ and $b$
Work Step by Step
Let $f(x) = sin~x$
Then $f'(x) = cos~x$
Note that $-1 \leq f'(x) \leq 1$ for all real numbers.
1. $f$ is continuous for all real numbers.
2. $f$ is differentiable for all real numbers.
$f$ satisfies the hypotheses of the Mean Value Theorem for all real numbers.
Choose any real numbers $a$ and $b$ where $a \lt b$
According to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$
Then:
$-1 \leq f'(c) \leq 1$
$-1 \leq \frac{f(b)-f(a)}{b-a} \leq 1$
$-1 \leq \frac{sin~b-sin~a}{b-a} \leq 1$
$\frac{\vert sin~b-sin~a \vert}{\vert b-a \vert} \leq 1$
$\vert sin~b-sin~a \vert \leq \vert b-a \vert$
$\vert sin~a-sin~b \vert \leq \vert a-b \vert$
Therefore, since $a$ and $b$ were chosen arbitrarily,
$\vert sin~a-sin~b \vert \leq \vert a-b \vert~~~~$ for all $a$ and $b$
