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 ;)!)
A scheme $X$ is qcqs (short for quasi-compact and quasi-separated) if and only if there exists a finite open affine cover $\{U_1,\cdots,U_n\}$ such that each intersection $U_i \cap U_j$ admits a finite open affine cover $\{V_{ij1},\cdots,V_{ijk_{ij}}\}$ (where $k_{ij} \in \mathbb N$ depends on $i$ and $j$).
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,
For (ii) we need to show for $\pi^{-1}(\operatorname{Spec} B)=\operatorname{Spec} A$, $A[g_i^{-1}]$ is a finite $B[g_i^{-1}]$-algebra for all $i$, implies that , $A$ is a finite $B$-algebra.
Let $A[g_i^{-1}]$ is generated by $\{f_{i1},\dots, f_{in}\}$ as $B[g_i^{-1}]$-module where $f_{ij}\in A$.
Consider $$\phi: B^{\oplus N}\to A$$ sending $$e_{ij}\mapsto f_{ij}.$$
Consider the cokernel $C$ of the map $\phi$. Now the $B$-module $C$ has the property that $C[g_i^{-1}]$ =0. Since $g_i$'s generate the unit ideal, we can conclude $C=0$.
Added: To see $C=0$, let $c\in C$. Then for each $i$, since $c\in C[g_i^{-1}]=0$, we get $g_i^mc=0$ for some $m$. Now $g_i$'s generate unit ideal, implies $g_i^m$'s generate the unit ideal. So 1 is a $B$-linear combination of $g_i^m$'s. So $c=1\cdot c=\sum g_i^mh_ic=0$.
To put (i) and (ii) together: Using the 'Affine communication lemma', $\operatorname{Spec} B\subset Y$ satisfies the property $P$ if $\pi^{-1}(\operatorname{Spec} B)$ is the spectrum of a finite $B$-algebra.
So you are done.
Best Answer
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.