For $0 \le x < 1$ is $\displaystyle \lim_{n \rightarrow \infty} s_{n} (x) = 1$ but for x=1 is $\displaystyle \lim_{n \rightarrow \infty} s_{n} (x) = 0$, so that $s_{n} (x)$ conveges pointwise in [0,1) but doesn't uniformly converge in [0,1)...
If \(\displaystyle f_n(x)\) is uniformally convergent then it is point-wise convergent. The difference is that uniform convergence is defined on sets.
Take for example the sequence of functions \(\displaystyle f_n(x)=x^n\) on the interval \(\displaystyle [0,1)\) this sequence is not uniformally convergent but any closed subset is. Essentially we can use the M-test to prove uniform convergence. Choose \(\displaystyle [0,b] \subset [0,1)\) then we have the following
\(\displaystyle x^n \leq b^n \,\,\, \forall \,\, x \in [0,b]\) since \(\displaystyle \lim b^n = 0 \) .\(\displaystyle f_n \) is uniformally convergent on \(\displaystyle [0,b]\).