Convergence of $f_n = \sin(x+\frac{1}{n})$ as $n \rightarrow \infty$ Is the convergence of $$f_n = \sin \left(x+\frac{1}{n} \right)$$ as $n \rightarrow \infty$, uniform, where $x \in \mathbb{R}$? 

It converges to $\sin x$ and $$||\sin (x+\frac{1}{n}) - \sin x ||_{\infty} \leq ||x + \frac{1}{n} - x||_{\infty} \rightarrow 0$$ as $n\rightarrow \infty$, so the convergence is uniform. 

Is this an okay argument?
Your Argument is Okay as well

since 

$$\sin a -\sin b  = 2\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})$$
Together with $$|\sin x|\le |x|$$

One get 

$$|\sin a -\sin b|  = |\sin(\frac{a-b}{2})\cos(\frac{a+b}{2})|\le |a-b|$$

Therefore 

$$\sup_{x\in \Bbb R}\|\sin (x+\frac{1}{n}) - \sin x |\leq \sup_{x\in \Bbb R}|x + \frac{1}{n} - x|  =\frac1n\rightarrow 0$$
I don't think this argument is okay, because the second displayed equation is hardly different from a restatement of what you are trying to prove.  Why is it true, or rather, what about the $\sin$ function makes it true? 

A good answer would make this clear, and would also make it clear if the same argument would work for any continuous periodic function, and not just the $\sin$ function.