Is a set contained in its Cartesian product with itself? Given the Cartesian product $X = A \times A$, is $A \subset X$?
If $A=\emptyset$, then indeed $A\subset X$. Otherwise, there is a copy of $A$ inside $X$ for every element of $A$, but the set $A$ itself is not necessarily exactly a subset of $X$. For every $a\in A$ we have $A\times\{a\}\subset X$ and $\{a\}\times A\subset X$. There is an obvious bijection between $A$ and $A\times\{a\}$.

A simple example for which it does not hold is $A=\{a\}$ with $a\neq\emptyset$. Then $X=\{(a,a)\}$ and $a\neq(a,a)$, so $A\not\subset X$.
Unfortunately its not.
Let $m \in A$. Then , $(m,m) \in A \times A=X$.
Now, What is the definition of $A \subset B$? --> Its $x \in A \Longrightarrow x \in B$.

Let your assumption $A \subset X$ is true, then $x \in A \Longrightarrow x \in X$
But we know $m \in A$ and $(m,m) \in X$. So this becomes $m \in A \Longrightarrow m \in X$. But we know $(m,m) \in X$ so this is clear contradiction. Hence your assumption is wrong. 
Here I don't used Quantifiers, because it needs a bit higher knowledge of logic, and you are saying I am not very good at set theory, so...