Solution check: Let $p:X\rightarrow Y$ be a closed continuous surjective map. Show that if $X$ is normal, then so is $Y$.

The argument seems correct, apart from a few slips:

1. you should start with $A\cap B=\emptyset$;
2. 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

1. $p^{-1}(A)\subseteq U$
2. $p^{-1}(B)\subseteq V$
3. $U\cap V=\emptyset$

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, $B\subseteq V_1$.