How to prove that $|\ln(2+\sin(x)) - \ln(2+\sin(y))| <= |x-y| \space \forall \space x,y \in \mathbb R$ Question states
Prove that for all $x$ and $y$ $\in R$, the following inequality is true:
$\lvert \ln(2+\sin(x)) - \ln(2+\sin(y))\rvert \le \lvert x-y\rvert$
i've gotten to the point that
$\frac{y-x}{2+\sin(c)}  = \ln\frac{2+\sin(y)}{2+\sin(x)}$  (y-x divided by 2+sin(c) is my f dash c from the mean value theorem
I asked my teacher that i should use mean value theorem here so please don't use anything other than this, but i have no idea how to push this problem further.
Also this is my first post so i'm sorry for the f dash (c) thing, mathjax is hard


I thought it might be instructive to present a way forward that relies on only elementary, pre-calculus tools.  To that end we proceed.


First, note that $\log(2+\sin(x))-\log(2+\sin(y))=\log\left(\frac{2+\sin(x)}{2+\sin(y)}\right)$.  
Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the iequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{t-1}{t}\le \log(t)\le t-1}$$
for all $t>0$.  Hence, with $t=\frac{2+\sin(x)}{2+\sin(y)}$, we have 
$$\frac{\sin(x)-\sin(y)}{2+\sin(x)}\le \log(2+\sin(x))-\log(2+\sin(y))\le \frac{\sin(x)-\sin(y)}{2+\sin(y)}\tag1$$

It is evident from $(1)$ that 
$$\left|\log(2+\sin(x))-\log(2+\sin(y))\right|\le |\sin(x)-\sin(y)|\tag2$$

Next, using the Prosthaphaeresis Identity, $\sin(x)-\sin(y)=2\cos\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$ along with $|\cos(\theta)|\le 1$ and $|\sin(\theta)|\le \theta$ reveals
$$|\sin(x)-\sin(y)|\le |x-y|\tag 3$$

Finally, using $(3)$ in $(2)$, we obtain the coveted inequality
$$\bbox[5px,border:2px solid #C0A000]{\left|\log(2+\sin(x))-\log(2+\sin(y))\right|\le |x-y|}$$
