Here's the definition:
A sequence of functions {$f_n$} is said to converge uniformly to $f$ on a set $S$ if, for every $\epsilon > 0$, there exists an $N$ (depending only on $\epsilon$) such that $n > N$ implies $\left|f_{n}\left(x\right)-f\left(x\right)\right|< \epsilon$ for every $x$ in $S$.
I don't see how this is different from usual convergence (in which $N$ depends on $\epsilon$ and $x$) for the following reason:
Since the sequence {$f_n$} converges pointwise to $f$, for every $x_j$ in $S$ and a given $\epsilon$, there exists an integer $N_j$ such that $\left|f_{n}\left(x_j\right)-f\left(x_j\right)\right|< \epsilon$ whenever $n > N_j$. This gives a set of integers $N_j$. Now if we pick $n$ greater than the greatest integer among the $N_j$'s, we will have $\left|f_{n}\left(x_j\right)-f\left(x_j\right)\right|< \epsilon$ for all $j$ and so, we can associate to every $\epsilon$ an $N$ which works for all $x$ in $S$ which is the definition of uniform convergence. What did I miss?
If you have an infinite set of positive integers there is no integer greater than the greatest among them.