Uniform convergence on singleton First, recall the definition of uniform convergence:
Consider functions $f_{n}:A\rightarrow\mathbb{R}$. The sequence of functions $f_{n}$ converges uniformly on set A to limit function f if $\lim\underset{x\in A}{\sup}|f_{n}\left(x\right)-f\left(x\right)|=0$
I see no restrictions on what A can be. But if I let A be singleton, then there will be no difference between pointwise and uniform convergences, so $f:\{1\}\rightarrow\mathbb{R}$ such that $f\left(x\right)=x^{n}$ converges uniformly on set $\{1\}$.
Second, recall the theorem that states: $f_{n}$ converges uniformly on $S_{1}$ and converges uniformly on $S_{2}$, then $f_{n}$ converges uniformly on $S_{1}\cup S_{2}$.
Now, let $S_{1}=[0,1),S_{2}=\{1\},f_{n}=x^{n}$. Hence, $x^{n}$ converges uniformly on $[0,1]$, which is wrong because it's well-known that uniform convergence holds only on $[0,1)$ for that function.
What went wrong? 

To answer your question "what went wrong":
The series of functions $f_n$, defined by $f_n(x)=x^n$, does not converge uniformly on $[0,1)$.
