Proof on Intersection of Sets I could not understand this property:
NOTE: P(A) is the Power Set.
$$P(A\cap B) = P(A)\cap P(B)$$
$$PROOF (\text{ Just the first part})$$
Consider an element say $a\in P(A\cap B)$. Then,
$$a\in P(A\cap B)$$
$$\Longrightarrow a\subseteq A\cap B$$
$$a\subseteq A \text{ and }a\subseteq B$$
$$a\in P(A) \text{ and }a\in P(B)$$
$$a\in P(A)\cap P(B)$$
$$\Longrightarrow P(A\cap B)\subseteq P(A)\cap P(B)$$
Could somebody explain the last step: how we get $P(A\cap B)\subseteq P(A)\cap P(B)$? How does the sequence of steps mean that $P(A\cap B)\subseteq P(A)\cap P(B)$? How do we get $P(A\cap B)$ as a subset?
 A: You started the proof with the assumption that
$$a\in P(A\cap B)$$
and ended with the conclusion that
$$a\in P(A)\cap P(B)$$
You have thus shown that there is an implication
$$a\in P(A\cap B)\to a\in P(A)\cap P(B)$$
which just means, since $a$ is arbitrary, that any member of the set $P(A\cap B)$ must also be a member of the set $P(A)\cap P(B)$: this is precisely the definition of a subset. Therefore we can conclude:
$$P(A\cap B)\subseteq P(A)\cap P(B)$$
A: In general to show a set $S$ is contained in another set $T$ you take any element $s\in S$ and show $s\in T$.  Therefore any element in $S$ is in $T$.  It must be then that $S\subseteq T$.  That's the logic of that step in the proof. 
A: If you show that every member of $X$ is a member of $Y$, you conclude that $X\subset Y$.  That's what "$\subset$" means.
Postscript inspired by a comment below:
It is not true that if just one element of $S$ is an element of $T$ then $S\subset T$.  But consider this argument.  "Let $s$ be a member of $S$.  Then we can show by the following argument that $s$ is purple."  Here you're proving that $s$ is purple by using no information about $s$ except that $s\in S$.  If you can do that, then it follows that every member of $s$ is purple.
