The argument seems correct, apart from a few slips:
- you should start with $A\cap B=\emptyset$;
- you seem to be confusing $\emptyset$ with $\{\emptyset\}$.
I'd make some passages clearer, in particular for showing where surjectivity is used.
Let $A$ and $B$ be disjoint closed subsets of $Y$. Then $p^{-1}(A)$ and $p^{-1}(B)$ are closed (by continuity of $p$) and disjoint (by general property of maps) subsets of $X$.
Since $X$ is normal, there are open sets $U$ and $V$ such that
- $p^{-1}(A)\subseteq U$
- $p^{-1}(B)\subseteq V$
- $U\cap V=\emptyset$
Since $p$ is closed, $p(X\setminus U)$ and $p(X\setminus V)$ are closed in $Y$. Let $U_1=Y\setminus p(X\setminus U)$ and $V_1=Y\setminus p(X\setminus V)$, which are open in $Y$.
Then
$$
U_1\cap V_1=Y\setminus(p(X\setminus V)\cup p(X\setminus U))
$$
Let's see that $p(X\setminus V)\cup p(X\setminus U)=Y$. If $y\in Y$, then $y=f(x)$ for some $x\in X$. Since $U\cap V=\emptyset$, we have either $x\in X\setminus U$ or $x\in X\setminus V$; so the thesis follows.
Therefore $U_1\cap V_1=\emptyset$.
Let $y\in A$. Suppose $y\notin U_1$; then $y\in p(X\setminus U)$, so $y=f(x)$ for some $x\in X\setminus U$. But this is impossible, because $x\in p^{-1}(A)\subseteq U$. Therefore $y\in U_1$.
Similarly, $B\subseteq V_1$.
An open set need not be of the form $U_1 \times V_1$, you can only say that $U = \cup_i (U_i \times V_i) - (A \times B)$, $U_i, V_i, i \in I$ are open in $X$, rep. $Y$. This doesn't help you for the proof, really.
A more direct appraoch is to write this difference $(X\times Y)- (A \times B)$ as a union of sets like $\{x\} \times Y, X \times \{y\}$, which are connected subspaces, and in such a way that there are intersections to ensure the connectedness of the union.
Best Answer
If you pick an element $u\in U_1\setminus U_2$ and $v\in V_2\setminus V_1$, it is clear that $(u,v)$ is in $(U_1\cup U_2)\times(V_1\cup V_2)$ but not in $(U_1\times V_1)\cup(U_2\times V_2)$.
Any union of sets of the basis of a topology is an open set, so this is a direct consequence of the definition of product topology.
A point $(u,v)$ is in $(U_1\times V_1)\ \cap (U_2\times V_2)$ iff it is in $U_1\times V_1$ and in $U_2\times V_2$, that is, if $u$ is in $U_1$ and in $U_2$ and $v$ is in $V_1$ and $V_2$. This is equivalent to saying that $u\in U_1\cap U_2$ and $v\in V_1\cap V_2$, which means that $(u,v)$ is in $(U_1\cap U_2)\times (V_1\cap V_2)$.
You can deduce all these laws regarding products, unions and intersections using just the definitions of these concepts and logic. The only logic laws that have been implicitly used in this reasoning are associativity and commutativity of conjunctions.