Proof.
The second statement follows from the first by elementary topology. According to Schemes, Lemma 26.18.5 to prove the first statement we may replace X by X_ s and S by \mathop{\mathrm{Spec}}(\kappa (s)). Thus we may assume that S = \mathop{\mathrm{Spec}}(k) is the spectrum of a field. In this case, let \mathop{\mathrm{Spec}}(A) = U \subset X be any affine open containing x. The point x corresponds to a prime ideal \mathfrak q \subset A such that \kappa (\mathfrak q)/k is an algebraic field extension. By Algebra, Lemma 10.35.9 we see that \mathfrak q is a maximal ideal, i.e., x \in U is a closed point. Since the affine opens form a basis of the topology of X we conclude that \{ x\} is closed.
\square
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