Showing $A \subset B \iff A\cap B=A$ Showing 
$A \subset B \iff A\cap B=A$
How would I show this?
My proof
Assume
i. $A \cap B \subset A$
ii.$A \subset A \cap B$
Let $x$ be any element.
Assume $x \in A \cap B$. Then $x \in A$ and $ x \in B$. By hypthesis $x \in A \rightarrow x \in B$ Thus $x \in A$
ii.
Let $ x \in A$ By hypthesis $x \in A \rightarrow x \in B$ 
thus $x \in A \cap B$.
Part 2
$ A\cap B=A \rightarrow A \subset B$
But I find myself stuck here.
 A: I think there is a cleaner way to write it. We want to prove $A \subset B \iff A \cap B  = A$
i) $A \subset B \implies A \cap B = A$
Here, we have to prove a set equality, so we have to prove both inclusions, assuming that $A \subset B$. First, let's prove that $A \cap B \subset A$. Let $x \in A \cap B$. Then $x \in A$ and $x \in B$. Well, $x \in A$, and we have $A \cap B \subset A$.
Now, we must get the other inclusion, $A \subset A \cap B$. Let $x \in A$. Since $A \subset B$, we have $x \in B$. But $x \in A$ and $x \in B$ means that $x \in A \cap B$, hence $A \subset A \cap B$.
Both inclusions give $A  = A \cap B$.
ii) $A \cap B = A \implies A \subset B$
Now, forget everything we've done until now. Assume that $A \cap B = A$, and we will prove that $A \subset B$. Let $x \in A$. That means, by hypothesis, that $x \in A \cap B$. So, $x \in B$, also, and we got $A \subset B \qquad\square$.
Notice where exactly I used every hypothesis, and the structure of the proof. If you still have any doubts, please say.
A: If $A \cap B = A$, then this means that for all $a\in A$, we have both $a \in A$ and $a \in B$ (definition of being in an intersection).
A: What you have written seems correct, but a bit confusing.
I might write the proof like this
Proof
We want to prove that 
$$
A \subset B \iff A\cap B=A
$$
Assume that the right hand side is true. That is, assume that $A\cap B = A$. We want to show that $A\subseteq B$. Let $a\in A$. Since $A = A\cap B$, we have $x\in A$ and $x\in B$. Hence $A\subseteq B$.
Assume now that the left hand side is true. That is, assume that $A\subseteq B$. We want to prove that $A\cap B = A$. We do this by proving to inclusions


*

*$A\cap B \subseteq A$ and

*$A\subseteq A\cap B$.


To show 1. let $x\in A\cap B$. Then $x\in A$ and $x\in B$. Hence $A\cap B \subseteq A$.
To how 2. let $x\in A$. Then since $A\subseteq B$ we have $x\in B$. So $x\in A$ and $x\in B$. Then by definition of intersection $x\in A\cap B$.
$\square$
A: *

*If $A \subset B$, then for every $a \in A$, we have $a \in B$ by definition of "$\subset$", thus $a \in A \cap B$. This shows that $A \subset A \cap B$. Now if $a \in A \cap B$ then $a \in A$ by definition of "$\cap$" and thus $A \cap B \subset A$. It follows that $A = A \cap B$.

*If $A \cap B = A$, then for every $a \in A$,  we have $a \in A \cap B$ by definition of "$=$" and thus $a \in B$ by definition of "$\cap$". This shows that $A \subset B$.
