Apply the Mean Value Theorem to show that:
$$|\cos(\frac{x_1}{2}) - \cos(\frac{x_2}{2})| \le \frac12|x_1 - x_2|,\ \text{ for any $x_1$ and $x_2$ in $[0, 2\pi]$}.$$
What I have done is this: use $f'(c) = [f(b) - f(a)]/(b-a)$. Here $2\pi$ is $b$ and 0 is $a$. I then treated both $x_1$ and $x_2$ as $c$ like this: $f'(x_1) = [f(2\pi) - f(0)]/2\pi. I found the function values at $2\pi$ and at $0$.
I got $[(\cos\pi - \cos\pi)]/(2\pi-0)$ which equals $0$. I then took the derivative with respect to $x_1$. So, $f'(x1) = -(1/2\sin(x_1/2) - \cos(x_2/2) = 0$.
I did the same thing again, this time using $x_2$ and $f'(x_2)$ and I got $0$ again.
So, by applying the MVT to $x_1$ and $x_2$ I get $0 \le (1/2)|x_1 - x_2|$.
I really do not think this process is correct.
Claim: If $0 \leq x_1 \lt x_2 \leq 2\pi$, then $$\left \vert \cos \frac{x_1}{2}- \cos \frac{ x_2}{2} \right \vert \leq \frac 12 \vert x_1 - x_2 \vert.$$
Proof: Define $f(x)= \cos \frac x2$. Then $f'(x)= - \frac 12 \sin \frac x2.$ Thus, for any $x_1, x_2 \in [0, 2\pi]$, the Mean Value Theorem (applied with $a=x_1, b= x_2$) tells us $\exists c$ with $x_1 \lt c \lt x_2$ such that
$$\left \vert \frac {f(x_1)-f(x_2)}{x_1-x_2} \right \vert \leq \vert f'(c) \vert.$$
Substituting, we have
$$\left \vert \frac{\cos \frac {x_1}{2} - \cos \frac{x_2}{2}}{x_1 - x_2} \right \vert \leq \frac 12 \vert \sin \frac c2 \vert \leq \frac 12.$$
The last inequality holds because $\vert \sin \frac c2 \vert \leq 1$ for any (real) value of $c$. Now just multiply through by $\vert x_1-x_2 \vert$ and the proof is complete.