Consider a partition of $A$ to $A_1,A_2,A_3,A_4$ such that $A_i=\lbrace 4k-i | 1 \leq k \leq s \rbrace$.By PHP,there exist an $i$ such that $|S \cap A_i| \geq \frac{s+1}{2}$,so there are three consecutive elements of $A_i$ which are in $S$,like $k,k+4,k+8$ but $k+(k+8)=2(k+4)$ and we are done.

If $2|s+1$ statement "so there are three consecutive elements of $A_i$ which are in $S$,like $k,k+4,k+8$" isn't true as you may take into $S$ such $\frac{s+1}{2}$ elements from $A_i$: $\lbrace 4\cdot 0+i,4\cdot 2+i,..., 4\cdot (s-1)+i\rbrace$ and none of them are three consecutive