Say $S_1 = \{a,b\}$ and $S_2=\{x,y,z\}$. Then
$$
S_1\times S_2= \big\{ (a,x),\ (a,y),\ (a,z),\ (b,x),\ (b,y),\ (b,z) \big\}.
$$
So you can look at it in either of two ways:
$$
\big\{ \underbrace{(a,x),\ (a,y),\ (a,z)},\ \underbrace{(b,x),\ (b,y),\ (b,z)} \big\}.
$$
$$
\big\{ \underbrace{(a,x),\ (b,x)},\ \underbrace{(a,y),\ (b,y)},\ \underbrace{(a,z),\ (b,z)} \big\}
$$
The first is a sum of two $3$s; the second is a sum of three $2$s. So the first is $2\times3$ and the second is $3\times 2$. That's one way of knowing that $2\times3$ is the same number as $3\times2$.
$$
|S_1|\times|S_2| = \sum_{a\in S_1}\ \sum_{x\in S_2}\ 1 = \sum_{(a,x)\in S_1\times S_2} \ 1.
$$
Best Answer
Take the set $D$ and make all possible pairs out of that set (since you are asked $D$ $\times$ $D$). In general, if $B$ is another set, and you want $D$ $\times$ $B$, then $D$ $\times$ $B$ = { (x,y) such that x $\epsilon$ $D$ and y $\epsilon$ $B$ } where if you replace $B$ by $D$, you get what you asked. Note that $D$ $\times$ $B$ and $B$ $\times$ $D$ aren't equal if $B$ is not he same as $D$. The first set (conventionally) comes on x-axis and the second on y-axis and so on if there are more than two sets.
Alternatively, for ease of understanding, fix one element of $D$ on x-axis and make pair of it with elements of $D$ on y-axis. This will give you points on top of the fixed point and then repeat it with another fixed element until you exhaust $D$ on x-axis. For example, fix point {2} from $D$ on x-axis and make pairs (2,y) where y varies in $D$ on y-axis. This will give you {2} $\times$ $D$. Now change the fixed point to another point in $D$ on x-axis and repeat the process until every element of $D$ is used as a fixed point, which wil give $D$ $\times$ $D$.