I am asking this because there is already a proof here when $X$ is quasi-compact, and it also seems to be necessary since almost everybody mentioning this result also mentions quasi-compactness (e.g. here), but I don't see how it is useful.
The proof would go as follows, let $X$ be any scheme, not necessarily quasi-compact and $Z \subset X$ a closed subset. Since a point of $Z$ is closed in $X$ iff it is closed in $Z$, it is enough to find a closed point in $Z$, which we can see as a closed subscheme of $X$, endowing it with the usual reduced subscheme structure. This time, $Z$ is not necessarily quasi-compact, but it can still be covered as a scheme with open affine subsets $(U_i)_{i \in I}$ with $I$ not necessarily finite. Let's now pick any point $p$ of $Z$ (we can since $Z \neq \emptyset$), it must be in one of the $U_i$, say $U_1$, and its closure necessarily lies in $U_1$ since $(\cup_{i \neq 1} U_i)^c \subset U_1$ is a closed set (which I can consider non-empty, otherwise $U_1$ would be superfluous in the covering and I could remove it). So writing $U_1=\operatorname{Spec}A_1$, $p$ is a prime ideal of $A_1$. If it is maximal, then its closure is itself and it is a closed point (in $U_1$ but also in $Z$). Otherwise, we can pick $p'$ corresponding to the maximal ideal containing $p$ and $p'$ is a closed point (in both $U_1$ and $Z$). In all cases we have found a closed point inside $Z$ without using quasi-compactness.
(this is essentially what I understood from the proof linked, but without using quasi-compactness, I can't see a flaw in it)
So is this proof correct? Else what am I missing? How is quasi-compactness relevant in this problem?
Best Answer
An open cover might in a non-compact topological space might not have an irredundant subcover. $\cup_{i \neq 1 \in I }U_i $ might always cover $U_1$.
Consider $\mathbb{R}$ with the open cover $(-n, n)_{n \in \mathbb{N}}$. Every subcover contains an open set that is in the union of the other open sets in the subcover.
A similar thing is going on in the paper I linked. In theorem 4.2, a scheme is constructed consisting of infinitely many points $\{p_n\}_{n \in \mathbb{N}}$, and the only open sets are of the form $\{p_0, ... p_n\}$.
Any open cover contains an element that is contained in the union of the others.