Proof for: Let A and B be sets s.t $ A \cap B = A $ iff $ A \subseteq B $ I am practicing some proofs involving sets and I would like to see if what I did was a valid proof because it seemed to be different from the one provided in the textbook I am using given that it did not use a contrapositive proof.
Claim: Let A and B be sets s.t $ A \cap B = A $ iff $ A \subseteq B $
Let $P(A,B): A \cap B = A$ and $q(A,B):  A \subseteq B $
1) Show that $ \neg q(A,B) \implies \neg p(A,B)$
2) Show that $q(A,B) \implies p(A,B)$
Proof: [1] Assume $A \not\subseteq B$ 
Thus, without loss of generality, assume that $x \in A$ and $ x \not\in B $
According to $ q(A,B): A\cap B \subseteq A \:\: \mathrm{and}\:\: A \subseteq A \cap B$ because, $ x \in A \:\: \mathrm{and}\:\: x\not\in B \implies x \in (A-B) \implies x \not\in A \cap B $ and $ x \in A $. Hence, $ A \cap B \not\subseteq A$. Because, $ x \in A $ and $ x \not\in A\cap B$, then $ A \not\subseteq A \cap B $. Hence, $ A \cap B \neq A $
[2]: Assume $ A \subseteq B $.
Let $ x \in A \implies x \in B $ without loss of generality.
Because $ x \in A $ and $ x \in B $, then $ x \in A \cap B $
$ \implies $ As $ x \in A \cap B $ and $ x \in A $, then $ A \cap B \subseteq A $.
Since, $ x \in A $ and $ x \in A \cap B $, then $ A \subseteq A \cap B $. Thus, $ A \cap B = A $. 
$\therefore A \cap B = A $ iff $ A \subseteq B $
Is what I did justified ? I would greatly appreciate any feedback or advice. 
Thank you

Everything is correct.
For the first part, or to be more precise, the first direction, you could also do with more direct way.
Suppose that $A\subseteq B$, therefore there is an element $x$ of $A$ which is in  $B$, thus $A\cap B\subseteq A$. But, $ A\subseteq A \cap B$. Therefore we have $A = A\cap B$.
