The slogan I remember is, "open immersions are stable under base change". Suppose you have a diagram
\[
\begin{matrix}
& & X' \\
& & \downarrow {\scriptstyle f} \\
U & \xrightarrow{i} & X
\end{matrix}
\]
and $i$ is an open immersion, in which case we might as well identify $U$ with an open subset of $X$. The fibered product exists and one realization of it is the open subscheme $f^{-1}(U)$ of $X'$. To show this you would just verify that it has the universal property. [You don't need to know that fibered products exist in general to do this.]
In the setting of the problem $X, X'$, and $U$ are affine, so you have another realization of $U \times_X X'$ as the spectrum of a ring. Hence there is an isomorphism between it and $f^{-1}(U)$.
A topological space is called quasi-separated if the intersection of any two quasi-compact open (not just any) subsets is quasi-compact. Check your definition.
Let $X$ be a scheme. Suppose the intersection of any two affine open subsets is a finite union of affine open subsets. Take two quasi-compact open subsets $U, V\subset X$. We want to show that $U\cap V$ is quasi-compact.
First prove the following easy lemmas:
Lemma 1 Every affine scheme is quasi-compact.
Lemma 2: If $Y$ is a topological space which is a finite union of quasi-compact spaces, then $Y$ is quasi-compact.
Lemma 3: Let $X$ be a scheme, then the affine open subsets form a base for Zariski topology.
Using the thrid lemma both $U,V$ are unions of affine opens. Using the fact that $U,V$ are quasi-compact, we find that $U,V$ are finite union of affine opens, say $U=\bigcup_{i=1}^n \mathrm{Spec}A_i$ and
$V=\bigcup_{j=1}^m \mathrm{Spec}B_j$. Then
$$
U\cap V= \bigcup_{i=1}^n \bigcup_{j=1}^m \mathrm{Spec}A_i\cap \mathrm{Spec}B_j $$
since the intersection of any two affine opens is a finite union of affine opens, then $U\cap V$ is a finite union of affine opens. Since every affine schme is quasi-compact (lemma 1), and since $U\cap V$ is a finite union of affine opens/quasi-compacts, then $U\cap V$ is quasi-compact (lemma 2).
Best Answer
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.