Every decidable language $L$ has an infinite decidable subset $S \subset L$ such that $L \setminus S$ is infinite
If $L$ has a finite alphabet, then $L$ is recursively enumerable.
Then, from such an enumeration $w_0,w_1,w_2,...$ of the words of $L$ you can take $S=\{w_0,w_2,w_4,...\}$, which will also be decidable. To check if a word $w$ is in $S$ check if it is in $L$. If it is then use the enumeration of $L$ to check if its position is even or not.