Is this proof valid for $A\subset B \implies A \cap B = A$?

Question

First semester for me. I need to prove the following:

If $A \subset B \implies A \cap B = A$.

I tried some simpler versions which don't quite satisfy me. So I tried the following:

Given $x \in A \cap B, A \subset B$ and $(A \cap B) \setminus A \neq \varnothing \implies \,\exists x \in (A \cap B) \setminus A : x \notin A \implies x \notin A \cap B$

Which is a contradiction. So $(A \cap B) \setminus A = \varnothing \implies A \cap B = A$

Answer 1

Let A $\subset$ B.

We show A $\cap$ B $\subset$ A. Suppose x$\in$ A $\cap$ B. By definition of intersection, x $\in$ A and x $\in$ B. So x $\in$ A. Thus A $\cap$ B $\subset$ A.

We now show A $\subset$ A $\cap$ B. Since A $\subset$ B by assumption, $\forall$ x $\in$ A, x $\in$ B. Thus $\forall$ x $\in$ A, x $\in$ A $\cap$ B. Thus A $\subset$ A $\cap$ B.

Therefore, A = A $\cap$ B.