Edited to add: Now that I've bullied man_in_green_shirt into choosing a definition of ordered pair, we can get to grips with this.
If an ordered pair $(a,b)$ is defined as $\{\{a\},\{a,b\}\}$, then we see that it has two elements (assuming that $a$ and $b$ are distinct), and therefore four subsets: $\emptyset, \{\{a\}\}, \{\{a,b\}\},$ and $\{\{a\},\{a,b\}\}$. Hence the four subsets of $(\{1,2\},\{4,6\})$ are:
$\emptyset$
$\{\{\{1,2\}\}\}$
$\{\{\{1,2\},\{4,6\}\}\}$
$\{\{\{1,2\}\},\{\{1,2\},\{4,6\}\}\}$
I hope I have that right. It looks gruesome in MathJax.
Original answer before the question was clarified:
The elements of $A$ are ordered pairs $(S_X,S_Y)$, where $S_X \subset X$ and $S_Y \subset Y$. So if $S_A$ is a subset of $A$, then the elements of $S_A$ are ordered pairs too. In particular, neither $\{1,4\}$ nor $\{1,2\}$ is a subset of $A$. Here is a typical subset of $A$:
$$\{(\{1,2\},\{4,6\}),(\emptyset,\{4,5,6\}),(\{2\},\{4,6\}),(\{1,2,3\},\{6\}),(\{1,3\},\{5\})\}$$