Why is $f_n(x) = x^n$ not uniformly convergent on $(0, 1)$?
Hint. You have $$\sup_{x\in(0,1)}f_n(x)=1.$$
Note that $\lim_{x\to1} f_n(x) = 1$ for all $n$. This breaks uniform convergence because we can get close enough to $1$ such that $f_N(x) > \frac12$ for any fixed $N$.
Hint If $x_n =1-\frac{1}{n}$ then $f_n(x_n) \to \frac{1}{e}$.
Use this to contradict $d(f_n(x_n), f(x_n))<\epsilon$.