A linear functional on is of the form , where is a bounded sequence of scalars. For f of that form, the condition tells you that for all k. You solve a recurrence relation of that type by using the auxiliary equation , which has the solutions . Then . In order for the sequence to be bounded, we must take the negative square root, getting . For that solution to be viable, we need two conditions. First, we must have (for the square root to be real). Second, we need (for the sequence to be bounded). That second condition reduces to .
Therefore if , we can conclude (as in the first part of the question) that and hence is a proper subspace. With a little care, you can reverse this argument to prove the converse: if is a proper subspace, then (by the Hahn–Banach theorem) there is a bounded linear functional which vanishes on . That functional must be of the form , where is a bounded sequence of scalars, and it follows as above that s must lie in the interval [–3,1].