Some questions on the inverse image of open set appeared in the fibre product of schemes. Suppose $X, Y, S$ be any scheme and $U \subseteq X$, $V \subseteq Y$ be open subsets. Let $p_1 : X \times_S Y \to X, p_2 : X \times_S Y \to Y$ be natural projections.
I've been finding the construction and the applications of fiber product of schemes and I've been curious why references treat followings harder than I expected.

$p_1^{-1}(U) = U \times_S Y$
$p_1^{-1}(U) \cap p_2^{-1}(V) = U \times_S V$

I ask this question because I'm not sure why we can't prove them just using set theoretical proof. It is obvious I can interpret them all as set.
For example, can't we just use
\begin{align}
p_1^{-1}(U) &= \{(x, y)/S \mid p_1(x, y) =x \in U \text{ and }x \sim_S y \}.\\
&= U \times_S Y
\end{align} directly?
Furthermore,
\begin{align}
p_1^{-1}(U) \cap p_2^{-1}(V) &= \{(x, y)/S \mid p_1(x, y) =x \in U \text{ and } p_2(x,y) = y \in V \text{ and }x \sim_S y \}.\\
&= U \times_S V
\end{align}
Why we should deal it with splitting two cases, $\subseteq$ and $\supseteq$?
I ADD CONTEXTS
The lemma where I messed up.
Actually, I met these issues in showing the diagonal map is locally closed.
I can't catch what is omitted in the above proof.
In particular,

Of course this morphism has image contained in the open $p^{-1}(V)∩q^{−1}(W)$. Thus $p^{-1}(V)∩q^{−1}(W)$ is a fibre product of $V$ and $W$ over $U$.

On the other hand, in Vakil, this is why I mentioned "splitting two cases",

What category they deal with? If it's not the category of sets, then what does $\subseteq, \supseteq$ mean? and in this case?
 A: CAUTION: The underlying set of the fiber product of schemes is NOT the fiber product of the underlying sets. Consider the fiber products $\operatorname{Spec}\Bbb C\times_{\operatorname{Spec}\Bbb C} \operatorname{Spec}\Bbb C$ versus $\operatorname{Spec}\Bbb C\times_{\operatorname{Spec}\Bbb R} \operatorname{Spec}\Bbb C$. The former is a single point, while the latter is $\operatorname{Spec} \Bbb C\otimes_{\Bbb R} \Bbb C$, two points! But the fiber product of the underlying sets in both cases is just a single point.
A correct description of the underlying set of the fiber product is as the quadruples $(x,y,s,\mathfrak{p})$ where $x\mapsto s$, $y\mapsto s$, and $\mathfrak{p}$ is a prime ideal of the tensor product $\kappa(x)\otimes_{\kappa(s)}\kappa(y)$ (ref Stacks 01JT). This explains why you should not work set-theoretically with just $(x,y,s)$ when dealing with the fiber product of schemes: it's not complete information determining the underlying set of $X\times_S Y$! 
Even if your set-theoretic attempt were enough, the text probably wants to talk about these results as schemes, not just as sets. In this case, there would need to be additional arguments made about the structure sheaf, which are missing from your attempt.

Responding to edit:
Let's deal with the $\subset$ and $\supset$ questions first. We are working in the category of schemes and we take these symbols to mean "is a subscheme of". For open subschemes, this reduces to the set-theoretic interpretation of "all points of the smaller one are points of the bigger one", since open immersions are isomorphisms on stalks. (Warning: this breaks very badly for closed subschemes. $V(x^2)\subset \Bbb A^1$ and $V(x)\subset \Bbb A^1$ have the same underlying points, $(x)$, but only the relation $V(x)\subset V(x^2)$ is true in the land of schemes.) So if we have two open subschemes $U_1,U_2$ of a scheme $X$, then $U_1\subset U_2$ as schemes iff $U_1\subset U_2$ as sets. Vakil uses two applications of this to see that $\delta^{-1}(U_{ij}\times_{V_i} U_{ij})=U_{ij}$.
The StacksProject proof is just the statement that the open subscheme $p^{-1}(V)\cap q^{-1}(W)$ satisfies the universal property of the fiber product $V\times_U W$: for any scheme $T$ with morphisms $f:T\to V$ and $g:T\to W$ so that the composites $T\to V\to U$ and $T\to W\to U$ agree, we get a unique morphism to $V\times_U W$. On the other hand, considering the relevant inclusions $V\to X$, $W\to Y$, and $U\to S$ we get a unique morphism $T\to X\times_S Y$. The image of this morphism $T\to X\times_S Y$ lands in $p^{-1}(V)$ by the assumption that the map $T\to X$ lands in $V$, and similarly for $T\to X\times_S Y$ landing in $q^{-1}(W)$ and the assumption $T\to Y$ lands in $W$.
So $T\to X\times_S Y$ lands in $p^{-1}(V)\cap q^{-1}(W)$: this means that given a pair of morphisms $T\to V$ and $T\to W$ so that the composites $T\to V\to U$ and $T\to W\to U$ agree, we have a unique morphism $T\to p^{-1}(V)\cap q^{-1}(W)$, which exactly shows that $p^{-1}(V)\cap q^{-1}(W)$ satisfies the same universal property as $V\times_U W$. But objects satisfying universal properties are unique up to unique isomorphism, so $p^{-1}(V)\cap q^{-1}(W)\cong V\times_U W$ canonically.
