I need to show that the sequence $\{\sin^{1/n}(x)\}$ converges uniformly on every closed proper subinterval of $[0,\pi]$. So let $a,b\in\mathbb{R}$ such that $0<a\leq x\leq b<\pi$. By definition I need to prove that for every $\varepsilon>0$ there is an integer $N$ such that if $n\geq N$, then $|\sin^{1/n}(x)-1|<\varepsilon$ for all $x\in [a,b]$.
I traied to see what happens with $M_{n}=\sup\limits_{x\in [a,b]}|sin^{1/n}(x)-1|$ but I can't find $N$ for the definition. I appreciate any hint
As indicated in the comments, a proper closed subinterval of $[0,\pi]$ may of the form $[0,\frac{\pi}{2}]$, in this case $\lim_n\sin^{1/n}(x)$ equals $0$ for $x=0$, and $1$ otherwise. Since the limit function is not continuous while $\sin^{1/n}(x)$ are, the convergence is not uniform.
But when $a>0$ and $b<\pi$, see that $M_n=\max\{|\sin^{1/n}(a)-1|,|\sin^{1/n}(b)-1|\}\to 0$, as $n\to\infty$.