Is $ (A∪B)×(C∩D)=(A∪C)×(B∪D)$ true for all sets A,B,C and D? Is $(A \cup B)\times (C ∩ D)=(A∪C)×(B∪D)$ true for all sets A,B,C and D?
I have no idea what to do here. How could I possibly go about trying to prove this? 
Or disprove it?
 A: You would go about trying to prove it exactly the way you go about trying to prove that two sets are equal: You fix an element in one, try showing it's in the other, and then switch.
To demonstrate, let's fix $(x,y)\in(A\cup B)\times(C\cap D)$. Then $x\in A\cup B$ and $y\in C\cap D$. In order to conclude $x\in (A\cup C)\times(B\cup D)$, we must have $x\in A\cup C$ and $y\in B\cup D$. While we do have $y\in D$ and hence $y\in B\cup D$, there doesn't seem to be a way to guarantee $x\in A\cup C$. At this point I suspect this inclusion is false. Let's find a counterexample.
The reason the above inclusion broke down is that $x\in A\cup B$ does not necessarily imply $x\in A\cup C$. Thus we want to find $A,B,C$, and $x$ such that $x\in B$ and $x\not\in A$ and $x\not\in C$. Let's take
$A=C=D=\{0\}$, $B=\{1\}$. Then
$$(A\cup B)\times (C\cap D) = \{0,1\}\times\{0\} = \{(0,0),(1,0)\}$$
and
$$
(A\cup C)\times (B\cup D) = \{0\}\times\{0,1\} = \{(0,0),(0,1)\}
$$
shows that $(A\cup B)\times(C\cap D)\not\subseteq (A\cup C)\times(B\cup D)$. In fact, this example demonstrates that the other inclusion fails as well.
A: Notice that $C \cap D$ can be empty even if $C$ and $D$ are not $\emptyset$, making the whole left hand site empty, but not the right hand side:
Take $A = B = C = \{1\}, D = \{2\}$
$$(A \cup B)\times (C \cap D) = \{1\}\times\emptyset = \emptyset$$
$$(A\cup C)\times (B\cup D) = \{1\}\times\{1,2\} = \{(1,1),(1,2)\}$$
They are obviously not equal.
