Clarification needed for proof of Theorem $17.2$ from Munkres Topology regarding necessary and sufficient conditions for a set to be closed. 
Without a picture, how do we know:
1.) $(X-C)\cap Y = Y-A$?
2.) $A = Y \cap (X-U)$?
Please help me understand these two. Thanks.
 A: For the first one, since $A = C \cap Y$, we are trying to see that
$$ (X - C) \cap Y = Y - (C \cap Y) $$
So say $x$ is an element of the left-hand set $(X - C) \cap Y$. 
Then $x \in (X - C)$ (which means $x$ is in $X$, but not in $C$) and $x \in Y$. 
So $x$ is in $Y$, but since $x$ is in not in $C$, it is not in $C \cap Y$. Thus $x$ is in $Y - (C \cap Y)$.
On the other hand, say $x$ is an element of the right hand set $Y - (C \cap Y)$. Then $x \in Y$, but $x$ is not an element of $C \cap Y$. $x$ therefore cannot be an element of $C$, since if it were in both $Y$ and $C$, it would be in $Y \cap C$. Also, since $Y \subseteq X$, $x$ is in $X$ as well as $Y$. We now have $x$ is in $X$ but not $C$, so $x \in X - C$ and $x \in Y$. Putting these together, we get $x \in (X - C) \cap Y$.
Thus, if $x$ is in one of these sets, it is in the other, so the two sets are equal.
For the second one, we are told that $Y - A$ is equal to the intersection of $U$ and $Y$
$$ Y - A = U \cap Y $$
So
$$ Y - (Y - A) = Y - (U \cap Y) $$
We see that the right hand side is the same as the right hand side of the first equation above replacing $C$ with $U$, so
$$ Y - (Y - A) = Y - (U \cap Y) = (X - U) \cap Y $$
$$ Y - (Y - A) = (X - U) \cap Y $$
And since $A$ is a subset of $Y$, $Y - A$ is the complement of $A$ in $Y$, and $Y - (Y - A)$ is the complement of the complement, which is just $A$ itself.
$$ A = (X - U) \cap Y $$
A: The identities are best remembered by writing out the meaning of the expressions.


*

*First note that, since $A=C\cap Y$ by assumption, we have $$Y\setminus A=Y\setminus(C\cap Y)=(Y\setminus C)\cup(Y\setminus Y)=(Y\setminus C)\tag{1}$$ by De Morgan's law. Now, 


$$\begin{split}(X\setminus C)\cap Y &=\{x\in X\text{ and }x\notin C\text{ and }x\in Y\} \\ &=\{x\in Y\text{ and }x\notin C\}\\ &=Y\setminus C \\ &=Y\setminus A,\end{split}$$ where the last equality follows from $(1)$.


*In this case, $Y\setminus A=U\cap Y$ by assumption. We also need an easy application of De Morgan's law first: $$A=Y\setminus(Y\setminus A)=Y\setminus(U\cap Y)=(Y\setminus U)\cup(Y\setminus Y)=(Y\setminus U).\tag{2}$$ Now, according to above, $$\begin{split} A &=\{x\in Y\text{ and }x\notin U\}\\ &=\{x\in Y\text{ and }x\in X\text{ and }x\notin U\}\\ &=\{x\in Y\text{ and }x\in X\setminus U\}\\ &=Y\cap(X\setminus U).\end{split}$$

A: We're given $A=C\cap Y$, so we need to show that $(X\setminus C)\cap Y = Y\setminus (C\cap Y)$.  De Morgan's Law says $Y\setminus (C\cap Y) = (Y\setminus C)\cup (Y\setminus Y) = Y\setminus C$. Likewise, $Y\cap (X\setminus C)$ is the set of all elements in $Y$ which are not in $C$, i.e. $Y\setminus C$.
Now for the second one.  We're given $Y\setminus A = U\cap Y$, and we want to show 
$Y\setminus (U\cap Y) = A = Y \cap (X\setminus U)$. But this is the same as what we showed above, with $C$ replaced with $U$.
A: Both of the claims follows directly from $A\cap (B-C)=(A\cap B)-(A\cap C)$ property of sets. Proof: let $x\in A\cap (B-C)$. So $x\in A$ $\wedge$ $x\in B$ $\wedge$ $x\notin C$ $\Rightarrow$ $x\in A\cap B$ $\wedge$ $ x\notin A\cap C$(since $x\notin C$) $\Rightarrow$ $x\in (A\cap B)-(A\cap C)$. Thus $A\cap (B-C)\subseteq (A\cap B)-(A\cap C)$. Conversely, let $x\in (A\cap B)-(A\cap C)$. So $x\in (A\cap B)$ $\wedge$ $x\notin (A\cap C)$ $\Rightarrow$ $x\in A$ $\wedge$ $x\in B$ $\wedge$ $x\notin C$(since $x\in A$ but $x\notin (A\cap C)$) $\Rightarrow$ $x\in A\cap (B-C)$. Thus $A\cap (B-C)\supseteq (A\cap B)-(A\cap C)$. So we have equality.
(1) $(X-C)\cap Y= (X\cap Y)-(Y\cap C)$. Since $Y\subseteq X$ and $Y\cap C=A$, we have $(X-C)\cap Y= Y-A$.
(2) $Y\cap (X-U)=(Y-X)-(Y\cap U)=Y-(Y\cap U)$. Since $(Y\cap U)=(Y-A)$, we have $Y\cap (X-U)=A$.
