Closed points of a scheme correspond to maximal ideals in the affines?

If $X=\mathrm{Spec}(A)$ is the spectrum of a discrete valuation ring, then the generic point $\eta$ is an affine open, namely $\mathrm{Spec}(k(\eta))$. Of course $\{\eta\}$ is closed in itself, but not in $\mathrm{Spec}(A)$. So this shows that a closed point in an affine open need not be closed in the ambient scheme.

This does, however, hold for schemes locally of finite type over a field $k$. Namely, if $X$ is such a $k$-scheme and $x\in X$ is closed in some affine open $\mathrm{Spec}(A)$, then $k(x)=A/\mathfrak{p}_x$ is an extension field of $k$ which is of finite type over $k$. So it is finite over $k$ by Zariski's lemma (also known sometimes as the Nullstellensatz). Now if $U=\mathrm{Spec}(B)$ is any affine open containing $x$, then $x$ corresponds to a prime $\mathfrak{p}_x^\prime$ of $B$, and $B/\mathfrak{p}_x^\prime$ is contained in $k(x)$ ($k(x)$ is the fraction field of this domain in fact). Since $k(x)/k$ is finite, we see that $B/\mathfrak{p}_x^\prime$ is a domain finite over a field, and so is itself a field, i.e., $\mathfrak{p}_x^\prime$ is a closed point of $\mathrm{Spec}(B)$. Since the affine opens cover $X$, it follows that $x$ is a closed point of $X$.

Perhaps there are other situations where closed points of affine opens are closed. I learned this particular fact from Qing Liu's book on algebraic geometry. If we're lucky maybe he will see this problem and add some further insight :)

A counter-example(the same as the Keenan Kidwell's) Let $A$ be a discrete valuation ring. Let $m$ be its unique maximal ideal. Let $f$ be a generator of $m$. Then $A_f$ is the field of fractions of $A$. Hence $x = {0}$ is a closed point of $D(f) = Spec(A_f)$. But it is not closed in $Spec(A)$.

Proposition Let $A$ be a Jacobson ring. Let $X$ be a scheme of locally finite type over $A$. Let $x$ be a closed point of an open subset $U$ of $X$. Then $x$ is a closed point of $X$.

Proof: Let $\bar {\{x\}}$ be the closure of $\{x\}$ in $X$. Let $y \in \bar {\{x\}}$. It suffices to prove that $x = y$. Let $V = Spec(B)$ be an affine open neighborhood of $y$. Then $x \in V \cap U$. There exists an affine open subset $D(f) = Spec(B_f)$ such that $x \in D(f) \subset V \cap U$, where $f \in B$. Since $x$ is closed in $U$, $x$ is closed in $D(f)$. Hence $x$ is a maximal ideal of $B_f$. Since $B_f$ is finitely generated over $A$ and $A$ is a Jacobson ring, $x$ is a maximal ideal of $B$. Hence $x$ is closed in $V$. Hence $\bar {\{x\}}\cap V = \{x\}$. Hence $x = y$ as desired. QED