which one of these proofs more acceptable for $A⊆B$ iff $A\cap B=A$? I've got two answers to prove that $A⊆B$ iff $A\cap B=A$
first,

(*) Assume that $A\cap B=A$. If $x\in A$ and $x\in B$ $\rightarrow $ $x\in (A\cap B)$. And if $y\in B$ but $y\notin A$ $\rightarrow $ $y\in (B-A)$. Therefore $x,y\in B$ and $x\in A$ but $y\notin A$, which shows that $A⊆B$.
(**) Assume that $A⊆B$. If $x,y\in B$ and $x\in A$ but $y\notin A$ $\rightarrow $ $x\in (A\cap B)$ and $y\in (B-A)$. Since $(A\cap B)$ has the same element as $A$, which is $x$, then $A\cap B=A$.

second,

(*) Assume that $A\cap B=A$. If $x\in A$ and $x\in B$ $\rightarrow$ $x\in (A\cap B)$. In particular, $x\in B$. This proves that $A⊆B$.
(**) Assume that $A⊆B$. If $x\in (A\cap B)$ $\rightarrow$ $x\in A$ and $x\in B$, so in particular $x\in A$. This proves that $A\cap B⊆A$. Now if $x\in A$ and $x\in B$ $\rightarrow$ $x\in (A\cap B)$. This proves that $A⊆A\cap B$. Together this implies that $A=A\cap B$.

Which one of these proofs more acceptable and why? Thank you so much for your time.
 A: Your first proof approach is dealing with two distinct arbitrary elements of (fill in the blank). This is not necessary or useful. It seems moreover that you misunderstand the purpose of an arbitrary element. (I will try to explain this further, but first, I will address your second attempt.)
Your second proof attempt avoids this issue, but still fails to complete either section of the proof. In (*), you have only proved that $A\cap B\subseteq B.$ In (**), you have proved that $A\cap B\subseteq A,$ but have failed to prove that $A\subseteq A\cap B.$

In general, one way to show that $C\subseteq D$ for some sets $C$ and $D$ is to show that for any $x,$ if $x\in C$, then $x\in D.$ Your second proof fell apart in assuming too much. Let me break it down for you.

(*) Assume that $A∩B=A.$

Good start. Now we need to suppose that $x$ is some element of $A$....

If $x∈A$  and $x∈B$

Oh, no! You just assumed that $x$ was an element of $A\cap B$! True, that happens to be equal to $A,$ but you're still skipping a step. Instead, we start by assuming $x\in A.$ Since $A=A\cap B$ by assumption, then we can conclude that...

$x∈(A∩B)$. In particular, $x∈B.$

Now we're done with that part.

(**) Assume that $A⊆B.$ If $x∈(A∩B)  →  x∈A$  and $x∈B$, so in particular $x∈A$. This proves that $A∩B⊆A$.

Doing fine so far. Now we need to prove that $A\subseteq A\cap B,$ so we start by assuming that $x$ is some element of $A$....

Now if $x∈A$  and $x∈B$

Again, we run into the same problem! Instead, assume only that $x\in A.$ Since $A\subseteq B,$ then we can conclude that $x\in B.$ So, since $x\in A$ and $x\in B,$ then....

$x∈(A∩B)$. This proves that $A⊆A∩B$. Together this implies that $A=A∩B$.


The reason for the arbitrary element is so that we don't have to make a different argument for every single element. (There could be infinitely many, so we might never get done!) Instead, we use a sort of "blanket argument" that applies to every element of the given set at once (even/especially if the given set is empty). There is really no need (or use) for the $y$ (that I can see). Perhaps you've seen a proof that used $x$ and $y$ in a way that seems similar to you. If this is the case, please let me know, and I'll try to make clear the distinctions and utility of such an approach.
A: For the first part, the first proof contains a completely irrelevant part. Who cares about that $y$? All you have to show is that if $x\in A$ then $x\in B$. There's nothing about any $y\in B-A$. 
The second proof is fairly okay, as Git Gud indicates in the comments, writing "$x\in A$ and $x\in B\to x\in A\cap B$" is not what you want to show. You want to take $x\in A$, and deduce that it is in $B$; for this you need to use the assumption that $A=A\cap B$, and therefore if $x\in A$ then $x\in A\cap B$, and so on.
The second part is also strange in the first proof. I'm not clear as to the role $y$ plays there as well. In addition, the fact that $A\cap B$ and $A$ have the same elements is exactly what you have to prove, you can't make an appeal to this.
In the second part's proof of the second part, in the second inclusion ($A\subseteq A\cap B$ part), you're not really proving what needs to be proved either. But it's much closer to an actual proof. You need to take an element from $A$ and show it is in $A\cap B$. You instead take an element in the intersection, then you claim it is in the intersection.
In short, the first proof of both part would have probably got no points from me; whereas the second proof would get about 50% of the grade in its current form.
A: Trim the fat.    You don't need to worry about elements not in $A$ but in $B$ (your '$y$').   You just need to worry about all of the elements that are in $A$ and whether or not they are in $B$ and/or the intersection.
I also prefer not mixing words and symbols as much as possible.   It leads to cleaner script.

Suppose A is a subset or equivalent of B.   This means every element of A is an element of B.   It follows that every element of A is an element of the intersection of A and B.   Thus A is the intersection of A and B.
Conversely, suppose A is the intersection of A and B.   This means every element of A is an element of B.   Thus A is a subset or equivalent of B.

Or symbolically:
$\begin{cases}\begin{align}
A\subseteq B & \iff \forall x\, (x\in A\rightarrow x\in B)
\\ & \iff \forall x\, (x\in A\leftrightarrow x\in A\land x\in B)
\\ & \iff \forall x\, (x\in A\leftrightarrow x\in A\cap B)
\\ & \iff A=A\cap B
\end{align}\end{cases}$
A: To be honest, I think you're writing a bit too much in your proofs, which is pretty normal for most people starting out in set theory. I was certainly no exception.

(*) Let $x\in A=A\cap B$. By definition of an intersection, this implies that $x\in B$. Since $x\in A \rightarrow x\in B$, we have shown that $A⊆B$.
(**) Assume that $A⊆B$. We must show that: $A⊆A\cap B$ and $A \cap B ⊆A$ to prove that $A = A\cap B$. Let $x \in A$; by the assmption, $x\in B$. Thus, $x \in A \cap B$. This proves that $A⊆A\cap B$. Now suppose that $x\in A \cap B$. This implies that $x\in A$. Thus, $x\in A\cap B \rightarrow x \in A$ implies $A\cap B $ ⊆ A.

That's it- you're done!
