Doing the limit we can see that in the open interval it converges pointwise to the constant function $f(x) = 0$. In the closed interval it doesn't converge uniformly because in $x=1$ $f(x) =1$ and when $0<x<1$ then $f(x)=0$. It isn't a continuous function although $f_n(x)$ is continuous for all $n$. So it can't be uniform.
However,let $0<ε<1$, we know the function increases as $x$ increases. So,
$|f_n(x) - f(x)| \leq ε^{n}$
And ε^(n) goes to 0 as n tends to infinity. Doesn't it mean that the convergence is uniform?
No, it doesn't mean that. Yes, $\lim_{n\to\infty}\varepsilon^n=0$, but that is not relevant here. Just note that$$(\forall n\in\mathbb N):\left(\sqrt[n]{\frac12}\right)^n=\frac12.$$But then $(f_n)_{n\in\mathbb N}$ doesn't converge uniformly to the null function. And, since it does converge pointwise to the null function, you can conclude that it doesn't converge uniformly at all.