Proof Verification: $\sin x$ is uniformly continuous on $\mathbb{R}$

Question

Proof: let $f(x)=\sin{x}$. Let $\epsilon>0$ be given. Choose $\delta := \epsilon$, then for $x,u \in \mathbb{R}$, $|x-u|<\delta$

$$\lvert f(x)-f(u)\rvert = \lvert\sin{x} - \sin{u}\rvert = \left\lvert 2\cos{\left(\frac{x+y}{2}\right)}\sin{\left(\frac{x-y}{2}\right)}\right\rvert<2\left\lvert\sin{\left(\frac{x-y}{2}\right)}\right\rvert<2\left\lvert \frac{x-y}{2}\right\rvert$$ $$=\lvert x-y\rvert <\delta=\epsilon.$$

Can anyone please verify this proof? Thank you.

Answer 1

It is almost perfect. The only slight problem lies in the inequality$$\left|2\cos\left(\frac{x+y}2\right)\sin\left(\frac{x-y}2\right)\right|<2\left|\sin\left(\frac{x-y}2\right)\right|.$$It should have been $\leqslant$ and not $<$.