Why is A a subset of P(A) only when A = nil? For example, if $A = \{1,2\}$ then $P(A) = \{\varnothing, \{1\}, \{2\}, \{1,2\}\}$
But then isn't $A \subseteq P(A)$ since all the elements that are in $A$ are also in $P(A)$? 
 A: No. $A$ is not a subset of $P(A)$, because $1 \in A$ but $1 \not\in P(A)$. We do have $\{1\} \in P(A)$, but that is something different than $1 \in P(A)$.
A: No. Not all of the elements in $A$ are in $P(A)$. Your error here is that you are assuming that $\{1\}$ means the same thing as $1$ and likewise for $2$, but that isn't true. $\{1\}$ represents the set with the number $1$ as its only element, and $1$ is the first natural number. It is true that
$$1\in \{1\}$$
But not
$$1\in \{\{1\}\}$$
However, you could say that
$$A\in P(A)$$
A: Note that $A \subseteq P(A)$ means every element in $A$ is also in $P(A)$. In your example, for example, $1$ is not an element of $P(A)$ because $P(A)$ is a group that contains groups and $1$ is not a group. $\{1\}$, however, is a group.
A: Let $A \subseteq P(A)$. Suppose $A \neq  \emptyset$, then by Axiom of Regularity,
$$\exists x \in A\text{ such that }x \cap A=\emptyset$$
As $A \subseteq P(A)$, therefore $$x \in P(A)$$ Also, $$\{x\} \in P(A)$$
Now, $x \in P(A)$ implies that $x \subseteq A$. Thus, $$x=x\cap A=\emptyset$$
Hence, $\emptyset \in A.$
Hence, we obtain the following $$A \subseteq P(A) \implies A=\emptyset \text{ or }\emptyset \in A$$
