Why $X\times\emptyset=\emptyset?$ If is true that $X\times\emptyset=\emptyset$, so take $A\times B$ with $U,V\subset A$ and $R,S\subset B$. We know that $(U\cup V)\times(R\cup S)=(U\times R)\cup (V\times S)$. So, take $X$ with $|X|\ge2$, then $X\times X=(X\setminus\{x\}\cup\{x\})\times(X\cup\emptyset)=(X\setminus\{x\}\times X)\cup(\{x\}\times\emptyset)=X\setminus\{x\}\times X?$. It's a paradox or I miss anything?
 A: $(U\cup V) \times (R\cup S) \ne (U\times R) \cup (V\times S)$.  It just doesn't.
Consider a $u\in U$ but $u \not \in V$ and $s \in S$ but $s \not \in R$.
then $u \in U\cup V$ and $s \in R\cup S$ so $(u,s )\in (U\cup V)\times (R\cup S)$
but $s\not \in R$ so $(u,s) \not \in U\times R$.  And $u \not \in V$ so $(u,s) \not \in V\times S$.  So $(u,s) \not \in (U\times R) \cup (V\times S)$ so $(U\cup V) \times (R\cup S) \ne (U\times R) \cup (V\times S)$.
.....
Any how $X \times \emptyset = \{(x,y)| x \in X; y \in \emptyset\}$.  But there are no $y \in \emptyset$ so there are no $(x,y) \in X\times \emptyset$.
So $X \times \emptyset$ is empty.
A: In general, $X \times Y$ is defined as the set that contains all pairs $(x,y)$ such that $x \in X$ and $y \in Y$. But, if $Y=\emptyset$, there are no such pairs! That is true since there is no $y$ such that $y \in \emptyset$. Therefore, according to the general definition, $X \times \emptyset = \emptyset$.
Another way to see that you want $X \times \emptyset$ to be the empty set is that you want $|X \times Y| = |X| \cdot |Y|$ to hold, where $|X|$ denotes the number of elements in set $X$. Let's focus on finite sets for now, although this is true for infinite sets as well. That is, in order to construct a pair $(x,y) \in X\times Y$, you have $|X|$ choices for $x$, and for every such choice, you have $|Y|$ choices for $y$. For $Y=\emptyset$, that tells you that you'd like $|X\times\emptyset|$ to be $0$, i.e. $X\times\emptyset = \emptyset$, since the empty set is the only set with $0$ elements.
Regarding the rest of your question, $(U\cup V)\times(R\cup S)=(U\times R)\cup (V\times S)$ isn't true. Consider any example with finite sets where $U \cap V = R \cap S = \emptyset$, and then simply count the size of each side. For the LHS we have:
$$|(U \cup V)\times(R \cup S)|= |U \cup V| \cdot |R \cup S| = (|U| + |V|) \cdot (|R| + |S|).$$
For the RHS:
$$|(U \times R) \cup (V \times S)| = |U\times R| + |V \times S| = |U|\cdot |R| + |V|\cdot |S|.$$
So, as long as none of $U,V,R,S$ is empty, the LHS can not be equal to the RHS.
Here is a concrete counterexample:
$$(\{0\}\cup\{1\})\times(\{a\}\cup\{b\})=\{0,1\}\times\{a,b\}=\{(0,a),(0,b),(1,a),(1,b)\}$$
whereas
$$\{0\}\times\{a\}\cup\{1\}\times\{b\}=\{(0,a),(1,b)\}.$$
