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.

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 $x\in A\cap B$.
$\square$
