Does it make sense to say that $\{1,4\}$ is a subset of $(\{1,2\},\{4,6\})$? Say we have $X=\{1,2,3\}$, $Y=\{4,5,6\}$ and take the cartesian product of their power sets $A=\mathcal{P}(X)\times \mathcal{P}(Y)$. Now, $(\{1,2\},\{4,6\})\in A$, although what are its subsets of this element? Is $\{1,4\}$ a subset of $A$, or is $\{1,2\}$?
It feels like $\{1,2\}$ should be a subset of $A$, although it wouldn't actually be in $A$, would it? $\{1,4\}$, on the other hand, would be an element of $A$.
Or does $(\{1,2\},\{4,6\})$ just have no proper subsets?
 A: (answer modified after question was edited)
$A$ and its subsets consist of ordered pairs of sets, not individual numbers, sets of numbers or sets of ordered pairs.
$(\{1,2\},\{4,6\})\in A.$ It is an ordered pair but not a set of ordered pairs. It is not a set, unless you are using a specific model for ordered pairs (which is sometimes done in logic handbooks where all objects must be sets).
Subsets are sets that contain zero or more elements from the given set and nothing else. So a subset of any subset of $A$ would automatically have to consist of ordered pairs.
A: 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 $A$:
$$\{(\{1,2\},\{4,6\}),(\emptyset,\{4,5,6\}),(\{2\},\{4,6\}),(\{1,2,3\},\{6\}),(\{1,3\},\{5\})\}$$
A: This may have to do with the quirky way of defining the ordered pair $(a,b)$ as the set $\{\{a\}, \{a,b\}\}$. Thus, $(\{1,2\}, \{4,6\} ) = \{ \{\{1,2\}\}, \{ \{1,2\}, \{4,6\} \} \}$.  Is $\{1,4\}$ a subset of that ? Not really. Can $\{1,4\}$ be a subset of some $(a,b) = \{ \{a\},  \{a,b\} \}$ ? Not really. So it seems the question is valid and has a negative answer. 
