Uniform continuity of a $ |\sin x| $ in $ [0,\infty) $ 
Given a function $ |\sin (x)| $ over the interval $ [0, + \infty) $ we have to check whether the function is uniformly continuous or not ?

So I begin by letting $ |x-y| < \delta $ then if I consider $ | |\sin x|-|\sin y| | $ then by triangular inequality we have $$ | |\sin x|-|\sin y| | < |\sin x - \sin y| = 2 |\sin(\frac {x-y}{2}) \cos (\frac {x+y}{2})| < 2|\sin(\frac {x-y}{2})|$$ 
Hence if I choose a $\epsilon > 2 \sin(\frac {\delta}{2} ) $ I can assure $ | |\sin x|-|\sin y| | < \epsilon$.
Is my proof correct ? i want some clarity on this ! also I am open to new methods or hints if you have any !!

In such a proof, you are given $\epsilon$ and need to choose $\delta$. It seems you’ve tried to do the opposite: you seem to be trying to choose a large enough $\epsilon$, when in reality $\epsilon$ should be fixed and your goal is to choose a small enough $\delta$. 
Regardless, an easier line of reasoning: by the mean value theorem, for any $x \neq y$ there is a $c$ between $x$ and $y$ with $$\lvert \sin(x) - \sin(y) \rvert = \lvert \cos(c) \rvert \lvert x-y\rvert \le \lvert x - y \rvert.$$ Do you see why this immediately implies uniform continuity in your case?
