Where does my proof of uniform continuity fail?

Question

I am trying to prove that

$f:R \to R f(x)=\sin x$

is uniformly continuous.

I have said:

Fix $\epsilon > 0$ and $\delta=\epsilon$

$|\sin x - \sin y| \le |\sin x| - |\sin y| \le 1 - 1 = 0 < \epsilon$

Answer 1

It fails because the correct triangle inequality is $|\sin x-\sin y|\le |\sin x|+|\sin y|$,

you must have "$+$" not "$-$"

Answer 2

$|sinx - siny| = |cosc(x-y)| = |cosc||sinx - siny| \leq |x-y|$. You can take it from here.

Answer 3

Hint: using a rather well known theorem

$$\exists\;c\in(x,y)\;\;s.t.\;\;|\sin x-\sin y|\le|x-y||\cos c|\le|x-y|$$

so....