Consider $f_n(x)=x^{1/n}$ on the interval $[0,1]$ . I know that $f_n$ converges pointwisely to the function $f$ defined by $f(x)=0$ when $x=0$ and $f(x)=1$ when $0<x \le 1.$
I am wondering whether there is a sub-interval of $[0,1]$ on which $f_n$ converges uniformly the constant function 1.
For any $n\geq 1$, $\lvert f_n-1\rvert = 1-f_n$ is decreasing on $[0,1]$ (check it by differentiation), so for any fixed $[a,1]$ with $a>0$ its maximum is attained at $f_n(a)$. It follows that, for any fixed $a\in(0,1]$, $$ \sup_{x\in[a,1]} \lvert f_n(x)-1\rvert = f_n(a) \xrightarrow[n\to\infty]{} 0 $$ the limit coming from pointwise convergence at $a$.
As you do not have uniform convergence on $[0,b]$ for any $b>0$ (as otherwise, the pointwise limit of the sequence of continuous functions $(f_n)_n$ would have to be continuous on $[0,b]$, and it is not), this is the best you can hope for.