How do I prove $U (A_1)\cap V (A_2) \neq \emptyset$? Can You help me to find where do we arrive contradiction? 
Let $(X,\mathscr T)$ be topological space with dense subset $D$ and  a
  closed,relatively discrete subset $C$ such that  $\mathscr{P}(D)\precsim$ $C.$ Then $(X,\mathscr T)$ is
  not normal.

Notations and definition in the theorem:-
$X\sim Y$- There is a bijection map from $X$ to $Y$.
$X\precsim Y-$ There is a subset $Y'$ of $Y$ such that $X \sim Y$
Relatively discrete.
A subset $A$ of a topological space $(X,\mathscr T)$ is relatively discrete provided that for each $a\in A$, there exists $U\in \mathscr T$ such that $U\cap A=\{x\}$.

I am having doubt in understanding the underlined statements. I have edited and made my doubts more precise.
Doubt 1:- How do I prove $C\setminus A$ is closed?
 My attempt:- It is enough to prove that $A$ is an open set. Let $x\in A\subset C$, then there is an open set $U\in \mathscr T: U\cap C=\{x\}.$ So, we can write $A=\bigcup_{x\in A}\{x\} $ is open in $C$. Hence, $A$ is closed in $C$. Hence Closed in $X$. Am I correct?
Doubt2:-How do I prove that $U(A_1) \cap V(A_2)\neq \emptyset$?
What is the idea of the proof afterwards? Where do we arrive at the contradiction?
 A: Regarding your first question: the earlier part of the first paragraph showed that if $B \subsetneq C$, then $B$ is closed in $C$. If we have a set $A \subsetneq C$ with $A \neq \varnothing$, then setting $B = A$ shows that $A$ is closed in $C$ (and hence closed in $X$), whereas setting $B = C-A$ shows that $C-A$ is closed in $C$ (and hence closed in $X$).
Regarding your second question: the assumption that $A_1 - A_2 \neq \varnothing$ implies that $A_1 \cap (C-A_2) \neq \varnothing$.  Since $A_1 \cap (C-A_2) \subset U(A_1) \cap V(A_2)$, this means that $U(A_1) \cap V(A_2) \neq \varnothing$.
Regarding the final contradiction: the second paragraph shows that there is an injection from $\mathcal{P}(C)$ into $\mathcal{P}(D)$.  If $\mathcal{P}(D) \subset C$, then $\mathcal{P}(C)$ would be in bijection with a subset of $C$, which would contradict Cantor’s theorem.
A: If $C$ is relatively discrete, this means that all $\{x\}$ where $x \in C$ are open in $C$ (because there is an $U \in \mathcal{T}$ such that $U \cap c=\{x\}$).
This implies, as all subsets are unions of singletons trivially, that all subsets of $C$ are open in $C$. As every subset of $C$ also has an open-in-$C$ complement (by the previous) it is in fact itself closed in $C$ and so closed in $X$.
Now by normality, we can find for every non-empty $A \subseteq C$ disjoint open subsets $U(A)$ and $V(A)$ such that $A \subseteq U(A)$ and $C-A \subseteq V(A)$, as $A$ and $C-A$ are disjoint closed subsets of $C$ and thus of $X$.
The aim is now to show that the function $f(A)=U(A) \cap D \in \mathscr{P}(D)$ from $\mathscr{P}(C)$ is 1-1 (injective)(you can add $f(\emptyset)=\emptyset$ to make it defined on the whole powerset, if you like). I'll rephrase the quoted argument slightly:
Suppose $f(A_1) = f(A_2) $ while $A_1 \neq A_2$.
$A_1 \neq A_2$ means either $A_1 - A_2 \neq \emptyset$ or $A_2 - A_1 \neq \emptyset$.
Say the former holds. Then we have some $p \in A_1, p \notin A_2$.
Then $p \in U(A_1) \cap V(A_2)$ by how the sets are chosen, and so we have a non-empty open set, which thus intersects the dense set $D$.
But $V(A_2) \cap f(A_1)= U(A_1) \cap D \cap V(A_2)\neq \emptyset$
while $V(A_2) \cap f(A_2)=V(A_2) \cap U(A_2) \cap D =\emptyset$ as $U(A_2)$ and $V(A_2)$ are disjoint. This contradicts $f(A_1)=f(A_2)$: the same set intersected with the same $V(A_2)$ cannot be both empty and non-empty. Contradiction 1.
We get another contradiction if $A_2 - A_1 \neq \emptyset$, using $V(A_1)$ intersections this time (check this, the text skipped this due to symmetry considerations, but I mention it for completeness).
These contradictions show that $f(A_1)=f(A_2)$ must imply $A_1=A_2$.
The last part is set theory: by Cantor's theorem we have that $\mathscr{P}(C)$ does not inject into $C$, or $\mathscr{P}(C) \not\precsim C$ .
But by the map $f$, $\mathscr{P}(C) \precsim \mathscr{P}(D) \precsim C$ (the last by the theorem's assumption). So combining: $\mathscr{P}(C) \precsim C$: a contradiction with Cantor's theorem.  QED
