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@abx I see your point. I meant to assume that the generic fibre $X_{K(S)}$ of $X\to S$ has a $K(S)$-rational point. This is stronger than the set $X(K(S))$ being non-empty (unless we consider only elements of $X(K(S))$ that correspond to the generic point Spec $K(S) \to S$. Then $X(K(S)) = X_{K(S)}(K(S))$ which can be certainly be empty (take a curve with no rational points over $\mathbb C(t)$).
