**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.
$\square$

## Comments (2)

Comment #2265 by yangan on

Comment #2294 by Johan on