The qc open subsets of $\mathrm{Spec}(A)$ have the form $\cup_i D(f_i)$ with finitely many $f_i \in A$. If we intersect two such sets, we optain $\cup_i D(f_i) \cap \cup_j D(g_j) = \cup_{ij} D(f_i g_j)$, which is a finite union of affine schemes, and therefore qc.
Okay, figured it out using some of the ideas in Matt E's post:
Suppose we have a cover of $Y$ by affine open subsets $Spec(U_{i})$ such that $\pi^{-1}(Spec(U_{i}))$ is quasiseparated. Given an arbitrary open subset $Spec(A)$ of $Y$, we wish to show that $\pi^{-1}(Spec(A))$ is quasiseparated. Using the affine communication lemma (See FOAG), we can find a cover of $Spec(A)$ by distinguished open subsets $D(f_{1}),\ldots,D(f_{n})$ such that $\pi^{-1}(D(f_{i}))$ is quasiseparated. Recall that an equivalent condition for a scheme $W$ to be quasiseparated is that it must be possible to cover the intersection of any two open affine subschemes with finitely many open affine subschemes. So let $Spec(B)$
and $Spec(C)$
be two such open subschemes of $\pi^{-1}(Spec(A))$
and note that the problem is reduced to the following special case:
$X$
is the union of two open affine subschemes $Spec(B)$
and $Spec(C)$
and $Y=Spec(A)$
. Let $D(f_{1}),\ldots,D(f_{n})$
be a finite cover of $Y$
such that $\pi^{-1}(D(f_{i}))$
is quasiseparated. We want to show that $X$
is quasiseperated.
Let $d_{C,i}$
denote the preimage of $D(f_{i})$
in $Spec(C)$
and $d_{B,i}$
denote the preimage in $Spec(B)$
. Then the $d_{B,i}$
and the $d_{C,i}$
form an affine cover of X
. Note that $d_{B,i}\cap d_{B,j}$
is the preimage of $D(f_{i})\cap D(f_{j})=D(f_{i}f_{j})$
in $Spec(B)$
, so it is affine. The analogous remark also applies to $d_{C,i}\cap d_{C,j}$
. Thus it remains to show that $d_{C,i}\cap d_{B,j}$
can be covered by finitely many affine open subsets: Note that $p\in d_{C,i}\cap d_{B,j}$
implies $p\in d_{B,i}$
since we must have $p\in Spec(B)$
and $\pi(p)\in D(f_{i}),D(f_{j})$
. Thus $d_{C,i}\cap d_{B,j}=d_{C,i}\cap(d_{B,j}\cap d_{B,i})$
We know that $d_{B,j}\cap d_{B,i}$
is a quasicompact (affine, in fact) open subset of $d_{B,i}\cup d_{C,i}$
, as is $d_{C,i}$
. Our original assumption tells us that $d_{B,i}\cup d_{C,i}$
is quasiseperated and it now follows that $d_{C,i}\cap d_{B,j}$
is quasicompact and can be covered by finitely many affine open sets. Thus, the space X
is quasiseparated.
Best Answer
I also wanted a topological proof, since it's nice to know what the proofs are in each language and knowing what the proof is when translated in another language (e.g. quasi-separatedness via diagonal morphisms) usually tells something a different way, which is not quite as enlightening. I cleaned up moji's proof (because I am not lazy ;)!)
Proof : ($\Rightarrow$) Pick a finite open affine cover $\{U_1,\cdots,U_n\}$ of $X$ by quasi-compactness. Affine schemes are qcqs, so the intersections $U_i \cap U_j$ are quasi-compact and therefore admit a finite open affine cover $\{V_{ij1},\cdots,V_{ijk_{ij}}\}$.
($\Leftarrow$) Let $U \subseteq X$ be a quasi-compact open subset. We claim that for each $\alpha=1,\cdots,n$, $U \cap U_{\alpha}$ is quasi-compact. It suffices to deal with the case of $\alpha=1$. Because $U$ is a scheme, its topology admits a basis consisting of quasi-compact open neighborhoods (take a finite open affine cover and the basis of distinguished open subsets of each of those affines). Write $$ U = \bigcup_{j=1}^n U \cap U_j = \bigcup_{j=1}^n \bigcup_{\ell \in L_j} W_{j\ell} $$ where $W_{j\ell} \subseteq U \cap U_j$ is a quasi-compact open subset. Since $U$ is quasi-compact, choose finite subsets $M_1 \subseteq L_1, \cdots, M_n \subseteq L_n$ such that the above equality still holds. Intersecting this with $U_1$, we get $$ U \cap U_1 = \bigcup_{j=1}^n \bigcup_{\ell \in M_j} W_{j\ell} \cap U_1. $$ Pick $j > 1$ and $\ell \in M_j$, so that for any $1 \le k \le k_{1j}$, the open subsets $V_{1jk}, W_{j\ell} \subseteq U_j$ are quasi-compact. Because $U_j$ is quasi-separated, $V_{1jk} \cap W_{j\ell}$ is quasi-compact. This means that $$ U \cap U_1 = \bigcup_{j=1}^n \bigcup_{\ell \in M_j} W_{j\ell} \cap U_1 \overset{(!)}= \bigcup_{j=1}^n \bigcup_{\ell \in M_j} W_{j\ell} \cap U_1 \cap U_j = \bigcup_{j=1}^n \bigcup_{\ell \in M_j} \bigcup_{k=1}^{k_{1j}} W_{j\ell} \cap V_{1jk} $$ is quasi-compact. (The $(!)$ is because $W_{j\ell} \subseteq U_j$ for each $j$. This seemed to be the cause of many incorrect edits to my proof.)
With this lemma in hand, if $U, U' \subseteq X$ are quasi-compact, then for $i=1,\cdots,n$, we see that $U \cap U' \cap U_i = (U \cap U_i) \cap (U' \cap U_i)$ is quasi-compact by the quasi-separatedness of $U_i$ and the quasi-compactness of $U \cap U_i$ and $U' \cap U_i$, so $X$ is quasi-separated.
Hope that helps,