This question is about showing that if $i: Y \longrightarrow X$ is a closed immersion of schemes with $X = \text{Spec }A$ affine, then $Y$ is affine. This question has had a lot of attention on this site already, particularly here and a question of my own here. I also wanted to ask about a different approach to this question, which is why I am hoping it's not considered a duplicate.
First of all I should state the definition of closed immersion I am working with
A morphism $i: Y \longrightarrow X$ is called a closed immersion if $i$ induces a homeomorphism between $Y$ and a closed subset of $X$, and the corresponding morphism of sheaves $i^{\#}: \mathcal{O}_{X} \longrightarrow i_{*}\mathcal{O}_{Y}$ is surjective.
I have also proved the following result independently of this exercise, which actually subsumes one of the hints given. In particular, I have proven the following
If $f: X \longrightarrow Y$ is an affine morphism, then for any affine open $\text{Spec }B \subseteq Y$, the preimage $f^{-1}(\text{Spec }B)$ is affine in $X$.
Having already proved this, my task is drastically reduced. Here is what I know so far, which seems to be the standard approach to the problem. Let $p \in Y$ be a point and let $\text{Spec }B$ be an affine neighbourhood of $p$ in $Y$. Then by definition of the subspace topology on $Y$, there is an open set $U$ of $X$ with $\text{Spec }B = Y \cap U$. Since $U$ is open in $X = \text{Spec A}$, we can find an $g \in A$ so that $p \in Y \cap D(g)$. Since $i^{-1}(D(g)) = Y \cap D(g)$, my entire task is now reduced to showing that $D(g) \cap Y$ is affine in $Y$, since then I can apply the above result on affine morphisms to see that $Y$ itself is affine.
This is where I am slightly confused. I have that $$ i|_{\text{Spec B}}: \text{Spec }B\longrightarrow \text{Spec A} $$ corresponds to a morphism of rings, say $$ \phi: A \longrightarrow B . $$ Now it seems "obvious" to me, that $D(g) \cap Y$ is given by the affine set $\text{Spec }B_{\phi(g)}$. It seems to me that this could be argued in the usual way, by taking the fibered product $$ \text{Spec }B \times_{\text{Spec }A} D(g) \simeq \text{Spec }\left( B \otimes_{A} A_{g} \right) \simeq \text{Spec }(B_{\phi(g)}). $$ My concern is whether this is rigorous in this context. I am just uncomfortable about the fact that we seem to have "cheated" by restricting $i$ to $\text{Spec B}$. Now the pullback isn't actually along $i$, but along $i|_{\text{Spec B}}$. Is this argument still valid here? Is that really all I have to say to show that $D(g) \cap Y$ is affine?
I also wanted to ask about how to see this in a potentially more general setting. Earlier in Hartshorne, in Exercise 2.16(a), we learn the following definition
If $X$ is a scheme and $f \in \Gamma(X, \mathcal{O}_{X})$ is a global section, then we define $$ X_{f} = \{ p \in X : f_{p} \not \in \mathfrak{m}_{p} \}. $$
The exercise there is to show that if $U = \text{Spec }B$ is affine in $X$, then $X_{f} \cap U = D\left( \tilde{f} \right)$, where $\tilde{f} \in \Gamma(\text{Spec }B, \mathcal{O}_{X})$ is the restriction of $f$ to $\text{Spec }B$. It was suggested to me that if we generalize this exercise to the case where $\text{Spec }B$ is not a subscheme of $X$, but rather just any affine scheme with a morphism $\text{Spec }B \longrightarrow X$, then this intersection can be viewed as a fibered product, and the result is much more useful for the exercise I was trying to do above. This seems at least reasonable to me, since if $U = \text{Spec }B \subseteq Y$ was my affine scheme of the closed subscheme $Y$ defined above, then wanting to show that $Y \cap D(g)$ is affine seems very reminiscent of this exercise, although I can't see the details of how this works. So my question can be summed up as follows
If $i: Y \longrightarrow X$ is our closed immersion defined above, then to what extend can the statement that $D(g) \cap Y$ is affine be regarded as a generalization of exercise 2.16(a) where instead of $\text{Spec }B \hookrightarrow X$ being an immersion, it is an arbitrary morphism?