To being with, I am using the following definition for an affine morphism of schemes.
A morphism $f: X \longrightarrow Y$ is said to be affine if there exists a cover $\{ \text{Spec }A_{i} \}$ of $Y$ by affines such that each $f^{-1}(\text{Spec }A_{i})$ is affine in $X$.
It is a well known result that if $f: X \longrightarrow Y$ is an affine morphism of schemes, then for any affine open set $\text{Spec }B \subseteq Y$, the preimage is affine. I am having a little trouble understanding the proof of this.
Without loss of generality, we may assume that $Y = \text{Spec }B$ is affine. Suppose $\{ g_{1}, g_{2}, \ldots , g_{m} \}$ generates the unit ideal in $B$, and suppose that $f^{-1}(D(g_{i})) = \text{Spec }A_{i}$ is affine in $X$. The part of the proof I am stuck on is showing that $X$ must then be affine. I have something of a proof, but I have been told this is not a good way to prove it, and that I am "missing the point" of the exercise. The standard procedure seems to be to use the following result
A scheme $X$ is affine if and only if there is a collection $\{ g_{1}, g_{2}, \ldots , g_{m}\}$ generating the unit ideal in $\Gamma(X, \mathcal{O}_{X})$, and the open sets $X_{g_{i}}$ are affine, where $X_{g_{i}}$ is the set of all points $p \in X$ such that the germ of $g_{i}$ at $p$ is not contained in the maximal ideal of the local ring $\mathcal{O}_{X, p}$.
With my situation as above, I have the morphism $f: X \longrightarrow \text{Spec }B$, which corresponds to a morphism of rings $$ \psi: B \longrightarrow \Gamma(X, \mathcal{O}_{X}), $$ and the $\psi(g_{i})$ generate the unit ideal in $\Gamma(X, \mathcal{O}_{X})$. My problem is that I can't see how to show that $X_{\psi(g_{i})}$ is affine in $X$. I'm expecting that the $X_{\psi(g_{i})}$ should be equal to the $\text{Spec }A_{i}$ but I can't see a nice way to show this.
I was also wondering if someone knows of a reference that doesn't do this in the language of $\mathcal{O}_{\text{Spec }B}$-algebras?