Closed Points of Quasi-Compact Schemes (Part 2)

Recall that quasi-compact schemes contain closed points- in fact, every closed subset of a quasi-compact scheme has closed point. In particular, every $x \in X$ is the generalization of some closed point, so any covering of the closed points must also be a cover of $X$. Thus to check if an open property holds for all points of $X$, it suffices to check if it holds for all closed points of $X$.

How about for properties $P$ that are not open?
Strategy 1.

Suppose P is  a property such that
for every $x$ not satisfying $P$, $\overline{\{x\}}$ contains another point  not satisfying $P$.
Then P holds for all $x \in X$ iff $P$ holds for all closed points of $X$.

Proof.
If this is true then we have a descending sequence
$$\overline{\{x_1\}} \supsetneq \overline{\{x_2\}} \supsetneq \overline{\{x_3\}} \supsetneq \overline{\{x_4\}} \overline \cdots.$$
On each affine piece, the sequence terminates as affine schemes are Noetherian. As $X$ as finitely many affine pieces, the sequence terminates in $X$ also. Thus there is $\overline{\{x\}}$ which contains no non-$P$ point other than $x$. Thus $x$ must be a closed point. (Contradiction)
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Strategy 2. 

Suppose $P$ is a property (of rings) such that if $A$ satisfies $P$ then $S^{-1}A$ satisfies $P$ (for every multiplicatively closed subset $S\subset A$ not containing $0$). Then $P$ holds for (the stalk of) all $x \in X$ iff $P$ holds for all closed points of $X$.

Proof.  Let $x \in X$. Then $\overline{\{x\}}$ contains a closed point $y$. Let $U= \text{Spec} A$ be an affine neighborhood of $y$. Then $U$ must contains $X$ as $y$ is in the closure of $x$. Thus we have $\mathbb{V}(\mathfrak{p}_x) \ni \mathfrak{p}_y$ so $\mathfrak{p}_y \supset \mathfrak{p}_x$ so we have a localization map
$$A_{\mathfrak{p}_y} \to A_{\mathfrak{p}_x}.$$
As $A_{\mathfrak{p}_y}$ has property $P$ and $P$ is compatible with localization, we have $A_{\mathfrak{p}_x}$ also has $P$.

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Example.

If $X$ is a quasi-compact space that is reduced at closed points, then it is reduced everywhere.

(Note: reducedness is a stalk local property. A scheme is reduced, i.e. its rings of sections are reduced, iff its stalks are all reduced).
Proof. Here we do not need to show that reduced-ness is an open condition (NOT true in general).  Rather we see that reducedness is compatible with localization (inverting elements cannot create nilpotents) so we can use strategy 2.