Let $A,B,C,D,X$ be sets.

1. $A\setminus B\subseteq A$ . This is obvious by definition.
2. If $A,B\subseteq X$ , then $$A\setminus B = A\cap B^\complement,\qquad (A\setminus B)^\complement = A^\complement \cup B,\qquad\mbox{and}\qquad A^\complement\setminus B^\complement = B\setminus A$$ where $^\complement$ denotes complement in $X$ .
   Proof. For the first equation, see here. The second equation comes from the first: $(A\setminus B)^\complement=(A\cap B^\complement)^\complement = (A^\complement)\cup (B^\complement)^\complement = A^\complement \cup B$ . The last equation also follows from the first: $A^\complement\setminus B^\complement = A^\complement \cap (B^\complement)^\complement = B\cap A^\complement = B\setminus A$ .
3. $A\subseteq B$ iff $A\setminus B=\emptyset$ .
   Proof. Since $A\subseteq B$ , $B^\complement \subseteq A^\complement$ . Then $A\setminus B= A\cap B^\complement \subseteq A\cap A^\complement=\emptyset$ . On the other hand, suppose $A\setminus B=\emptyset$ . Then $A\cap B^\complement = \emptyset$ by property 1, which means $A\subseteq (B^\complement)^\complement=B$ .
4. $A\cap B=\emptyset$ iff $A\setminus B=A$ .
   Proof. Suppose first that $A\cap B=\emptyset$ . If $a\in A$ , then $a\notin B$ , so $a\in A\setminus B$ , and hence $A\subseteq A\setminus B$ . The equality is shown by applying property 1. Next suppose $A\setminus B=A$ . If $a\in A$ , then $a\in A\setminus B$ , so $a\notin B$ , which means $A\subseteq B^\complement$ , or $A\cap B=\emptyset$ .
5. $A\setminus\emptyset = A$ and $A\setminus A = \emptyset = \emptyset\setminus A$ .
   Proof. The first equation follows from property 4 and the last two equations from property 3.
6. (de Morgan's laws on set difference): $$A\setminus (B\cap C)=(A\setminus B)\cup (A\setminus C)\qquad \mbox{ and }\qquad A\setminus (B\cup C) = (A\setminus B)\cap (A\setminus C).$$
   Proof. These laws follow from property 2 and the de Morgan's laws on set complement. For example, $A\setminus (B\cap C)=(A\setminus B)\cup (A\setminus C) = A\cap (B\cap C)^\complement = A\cap (B^\complement \cup C^\complement) = (A\cap B^\complement) \cup (A\cap C^\complement) = (A\setminus B)\cup (A\setminus C)$ . The other equation is proved similarly.
7. $A\setminus(A\cap B) = A\setminus B = (A\cup B)\setminus B$ .
   Proof. The first equation follows from property 6: $A\setminus (A\cap B)=(A\setminus A)\cup (A\setminus B)= A\setminus B$ by property 5. Next, $(A\cup B)\setminus B=(A\cup B)\cap B^\complement = (A\cap B^\complement)\cup (B\cap B^\complement)= A\cap B^\complement =A\setminus B$ , proving the second equation.
8. $(A\cap B)\setminus C=(A\setminus C)\cap (B\setminus C)$ .
   Proof. Using property 2, we get $(A\cap B)\setminus C=(A\cap B)\cap C^\complement = (A\cap C^\complement)\cap (B\cap C^\complement) = (A\setminus C)\cap (B\setminus C)$ .
9. $A\cap (B\setminus C)=(A\cap B)\setminus (A\cap C)$ .
   Proof. $(A\cap B)\setminus (A\cap C) = (A\cap B)\cap (A\cap C)^\complement = (A\cap B)\cap (A^\complement \cup C^\complement) = ((A\cap B)\cap A^\complement)\cup ((A\cap B)\cap C^\complement) = (A\cap B)\cap C^\complement = A\cap (B\cap C^\complement) = A\cap (B\setminus C)$ .
10. $(A\setminus B)\cap (C\setminus D) = (C\setminus B)\cap (A\setminus D)$
    Proof. Expanding the LHS, we get $A\cap B^\complement \cap C \cap D^\complement$ . Expanding the RHS, we get the same thing.
11. $(A\setminus B)\cap (C\setminus D) = (A\cap C)\setminus (B\cup D)$ .
    Proof. Starting from the RHS: $(A\cap C)\setminus (B\cup D)=((A\cap C)\setminus B)\cap ((A\cap C)\setminus D)=(A\setminus B)\cap (C\setminus B)\cap (A\setminus D)\cap (C\setminus D)=(A\setminus B)\cap (C\setminus D)$ , where the last equality comes from property 10.

Remarks.

1. Many of the proofs above use the properties of the set complement. Please see this link for more detail.
2. All of the properties of $\setminus$ on sets can be generalized to Boolean subtraction on Boolean algebras.