a proof by contradiction should work. what if $s_n \le 0$ for all $n$? consider the cases $s_n = 0$ for all $n$ and $s_n < 0$ for all $n$ separately
3. In the definition of sequence convergence, set $\varepsilon = \frac{s}{2}$.
Then $\left( {\exists N} \right)\left( {\forall n > N} \right)\left[ {\left| {s_n - s} \right| < \varepsilon } \right]\, \Rightarrow \,s_n > \frac{s}{2} > 0$