Using the Mean Value Theorem, prove that $|\sin{a} - \sin{b}| \leq |a - b|$ $\forall a, b \in \mathbb{R}$ 
Using the Mean Value Theorem, prove that $|\sin{a} - \sin{b}| \leq |a - b|$ $\forall a, b \in \mathbb{R}$.

I'm working towards figuring out an approach for finding that $|\sin{a} - \sin{b}| \leq |a - b|$ $\forall a, b \in \mathbb{R}$, but I've not yet included an application of the MVT, and I believe that my approach has some redundancy (or at least isn't that elegant). Furthermore, I'm not even so certain what I've written is at all very helpful in proving this conclusion.

$$
\begin{align*}
\\ \text{Assume } \forall \sin{x} \implies \sin{x} = \sin{(x \bmod 2\pi)}
\\ \text{case: }
    a &> b \wedge a \leq \pi \wedge b \leq \pi \implies
\\  1 &\geq \sin{a} \geq 0 \wedge  1 \geq \sin{b} \geq 0 \implies
\\  0 &\leq |\sin{a} - \sin{b}| \leq 1
\\
\\ \text{case: }
    a &> b \wedge a \geq \pi \wedge b \leq \pi \implies
\\ -1 &\leq \sin{a} \leq 0 \wedge  1 \geq \sin{b} \geq 0 \implies
\\  0 &\leq |\sin{a} - \sin{b}| \leq 2
\\
\\ \text{case: }
    a &> b \wedge a \geq \pi \wedge b \geq \pi \implies
\\ -1 &\leq \sin{a} \leq 0 \wedge -1 \leq \sin{b} \leq 0 \implies
\\  0 &\leq |\sin{a} - \sin{b}| \leq 1
\\
\\ \text{case: }
     a &= b \implies
\\   0 &= |\sin{a} - \sin{b}| = |a - b|
\\
\\ \text{case: }
    a &< b \wedge a \leq \pi \wedge b \leq \pi \implies
\\  1 &\geq \sin{a} \geq 0 \wedge  1 \geq \sin{b} \geq 0 \implies
\\  0 &\leq |\sin{a} - \sin{b}| \leq 1
\\
\\ \text{case: }
    a &< b \wedge a \leq \pi \wedge b \geq \pi \implies
\\  1 &\geq \sin{a} \geq 0 \wedge -1 \leq \sin{b} \leq 0 \implies
\\  0 &\leq |\sin{a} - \sin{b}| \leq 2
\\
\\ \text{case: } a &< b \wedge a \geq \pi \wedge b \geq \pi \implies
\\ -1 &\leq \sin{a} \leq 0 \wedge -1 \leq \sin{b} \leq 0 \implies
\\  0 &\leq |\sin{a} - \sin{b}| \leq 1
\end{align*}
$$

I find it's fairly intuitive for $|a - b| \geq 2$ that $|\sin{a} - \sin{b}| \leq 2$, considering $\sin{x} \leq 1 \text{ } \forall x \in \mathbb{R}$. But grasping and considering cases for $|a - b| < 2$ seems a little less intuitive, as it is perhaps conceivable (but not necessarily true) that $|\sin{a} - \sin{b}| > |a - b|$ for some values where $|a - b| < 2$.
Insight?

Edit:  I've refined my proof to the following structure:
The Mean Value Theorem states: a function $f$ which is continuous on the closed interval $[a, b]$ $^{\textbf{(1)}}$ and differentiable on the open interval $(a, b)$ $^{\textbf{(2)}}$ has at least one value $c: a < c < b$ where $f'(c) = \dfrac{f(b) - f(a)}{b - a}$.
Set $f(x) = \sin{x} \implies f(x)$ is continuous and differentiable $\forall x \in \mathbb{R}$ and all sub-intervals $^{\textbf{(1, 2)}}$ $ \therefore$ when $\exists a, b: b < c < a \implies \exists f'(c) = \dfrac{f(a) - f(b)}{a - b} \implies \cos{c} = \dfrac{\sin{a} - \sin{b}}{a - b}$. Take the absolute value of both sides of this equality to find $\dfrac{|\sin{a} - \sin{b}|}{|a - b|} = |\cos{c}|$, and since $\dfrac{|\sin{a} - \sin{b}|}{|a - b|} = \dfrac{|\sin{b} - \sin{a}|}{|b - a|}$, this holds true $\forall a, b \in \mathbb{R}$. Since $|\cos{x}| \leq 1 \text{ } \forall x \in \mathbb{R} \implies |\cos{c}| \leq 1 \implies \dfrac{|\sin{a} - \sin{b}|}{|a - b|} \leq 1.$ Multiplying across the inequality by $|a - b|$ finds the result: $|\sin{a} - \sin{b}|  \leq |a - b|$.

Hint: by MVT,
$$\frac{\sin a-\sin b}{a-b}=\cos c$$
for some $c$.  Now what do you know about the RHS?
