How do you prove that $A\cap B=B\Leftrightarrow B\subseteq A$? How do you prove that $A\cap B=B\Leftrightarrow B\subseteq A$? My thought process so far was as follows:
By the way, this is what the exercise looks like
When is $B\subseteq A$ ($B$ is a subset of $A$)?

if $B\subset A$  or if $B=A$

*

*$(B\subset A) \Leftrightarrow (x\in A \; \forall x\in B)$

*$(A\subset B) \Leftrightarrow (x\in B \; \forall x\in A)$

*$(A=B) \Leftrightarrow (A\subset B) \land (B\subset A) \Leftrightarrow (x\in A \;\forall x\in B) \land (x\in B \;\forall x\in A)$

*$(B=A) \Leftrightarrow (A=B)$

*$(B\subseteq A) \Leftrightarrow (B\subset A) \lor (B=A) \Leftrightarrow (x\in A \;\forall x\in B) \lor ((x\in A \;\forall x\in B) \land (x\in B \;\forall x\in A))$

and because in general $A \lor (A \land B) \Leftrightarrow A$:

$(x\in A\;\forall x\in B) \lor ((x\in A \;\forall x\in B) \land (x\in B \;\forall x\in A))\Leftrightarrow (x\in A\;\forall x\in B)$.

But this would mean that $B\subset A\Leftrightarrow B\subseteq A$. So I must have done something wrong here or misunderstood, I guess 

If I change it to $B\subsetneq A$ ($B$ is a proper subset of $A$):

*

*$(B\subsetneq A) \Leftrightarrow (x\in A\;\forall x\in B) \land (\exists x\in A : x\notin B)$

*$A\subsetneq B \Leftrightarrow (x\in B\;\forall x\in A) \land (\exists x\in B : x\notin A)$

*$A=B \Leftrightarrow (x\in A\;\forall x\in B) \land (x\in B\;\forall x\in A)$

*$B=A \Leftrightarrow A=B$

*$B\subseteq A \Leftrightarrow (B\subsetneq A) \lor (B=A) \Leftrightarrow ((x\in A\;\forall x\in B) \land (\exists x\in A : x\notin B)) \lor ((x\in A\;\forall x\in B) \land (x\in B\;\forall x\in A))$

and because in general $(A \land B) \lor (A \land C) \Leftrightarrow (B \lor C) \land A$ (because of distributivity)

$B\subseteq A \Leftrightarrow ((\exists x\in A : x\notin B) \lor (x\in B\;\forall x\in A)) ∧ (x\in A\;\forall x\in B)$

But is this even correct so far?

And now I looked at the other part, of which I don't even fully understand the meaning $(A\cap B)=B$:

I assume that you can write it like this

$x \in A\cap B = x \in B$

$\Leftrightarrow (x\in A \land x\in B) = x \in B$
And this is where I stopped, because I have no idea how to continue with this information, assuming it is even correct so far.
 A: Your conclusion that your first argument shows $B\subset A\iff B\subseteq A$ is incorrect.  Let $p=\forall x\in B: x\in A$, and let $q=\forall x\in A: x\in B$  Then the statement you have proved is $$p\lor(p\land q)\iff p$$ which is a tautology.
You seem to be going around in circles.  I suggest that you first draw a Venn diagram to see why the statement is true.  Once you understand that, you'll find that it's much easier to write a proof.
A: I wouldn't try to use equivalences. I would focus on one direction at a time, using implications.
$(\Longrightarrow):$ Suppose that $A \cap B = B$. We want to show that $B \subseteq A$. To this end, choose any $x \in B$. But since $B = A \cap B \subseteq A$, it follows that $x \in A$, as desired.
$(\Longleftarrow):$ Suppose that $B \subseteq A$. We want to show that $A \cap B = B$. It's easy to see that $A \cap B \subseteq B$, so it remains to show that $B \subseteq A \cap B$. To this end, choose any $x \in B$. Then since $B \subseteq A$, it follows that $x \in A$. But then $x \in A \cap B$, as desired.
A: $$
(A\ \cap\ B\, ) = B \iff \forall x:[(x\in A\ \land x\in B) \leftrightarrow x\in  B]
$$
$$
p := (x\in A),\qquad q := (x\in B)
$$
$$
(p \land q) \leftrightarrow q\\
\iff ((p \land q) \rightarrow q) \land ((p \land q) \leftarrow q) \\
\iff (\neg(p \land q) \lor q) \land ((p \land q) \lor\neg q) \\
\iff (\neg p  \lor \neg q  \lor q) \land (p \lor\neg q)\land (q \lor\neg q) \\
\iff (p \lor\neg q)\\
\iff (q \to p)\\
$$
$$
\forall x:[(x\in A\ \land x\in B) \leftrightarrow x\in  B]
\iff  \forall x: [x\in B \to x\in A] \\
(A\ \cap\ B\, ) = B \iff  B \subseteq A
$$
A: I think that what's happening is a confusion with the notation and the definitions.
I'm not sure why, but you seem to have the same definition for $B\subset A$ and $B\subseteq A$, but you want to think of $B\subset A$ as $B\subsetneq A$. I think this comes from the ambiguity of $B\subset A$, since some people think of it as $B\subseteq A$, but others prefer $B\subsetneq A$. I think you are thinking of it as the latter case, but you're using the definition of the former one.
To put it in words, I think you think $B\subset A$ means that every element of $B$ is in $A$ too, but $A$ and $B$ cannot be equal. But if they are not equal there has to be some element in one of them that isn't in the other; since every element of $B$ is in $A$, the element they don't share must be in $A$. So $B\subset A$ means that every element of $B$ is in $A$ and there is some element in $A$ that isn't in $B$. This is the same definition as $B\subsetneq A$, so for you they express the same thing. Expressing the definition in mathematical notation: $(x\in A\;\forall x\in B)\land(\exists x\in A : x\notin B)$.
The definition you have for $B\subset A$ is $x\in A\;\forall x\in B$. This, putting it in words, means that every element of $B$ is an element of $A$ too, but this allows the case when every element of $A$ is in $B$ too, since we are not restricting the definition for a particular case. What I want to express is that the fact that every element of $B$ is in $A$ is true when they are the same set, so $B\subset A$ would mean the same as $B\subseteq A$.
As I mentioned, there is no correct answer for which definition to use. You can take either of them; just make sure your conception of the meaning of $B\subset A$ and your mathematical definition for it agree. However, as I was pointed out in the comments, if you choose that $B\subset A$ is the same as $B\subsetneq A$ then your statement is not true, since $A\cap B=B\nRightarrow B\subsetneq A$. Thus, you're probably using the $B\subseteq A$ definition.
A: Wow.  You are worrying and trying to get caught up in notation way too much!
First of all, $B \subseteq A$ means that $B$ is a subset of $A$.   It allows the possibility that $B$ could be equal to $A$ but that still means $B$ is a subset of $A$ because all sets are subsets of themselves.
So saying "$B$ is a subset of $A$ or $B$ is equal to $A$" is not an exclusive OR such as  "$x$ is a mammal or $x$ is an insect".  It's not even an overlapping OR such as "$x$ is a mammal or $x$ is a carnivore".  It is a subordinate OR such as "$x$ is a mammal or $x$ is an antelope".
This is a case of $M$ or $N$ but where $N$ is only true if $M$ is.  In these cases $M$ or $N\iff M$.  Is a perfectly true statement.

"$B$ is a subset of $A$ or $B= A\iff B$ is subset of $A$"  is a perfectly TRUE statement.
Pf: $\implies$:  If $B$ is a subset of $A$ or $B=A$ then either i) $B$ is a subset of $A$.  If so we are done.  or ii) $B=A$.  If so then $A$ is subset of itself so $B$ which is $A$ is a subset of $A$ which is itself.  So $B$ is a subset of $A$ or $B= A\implies B$ is subset of $A$
$\Leftarrow$:  If $M$ is true then $M$ or $X$ is true for any statement $X$ so if $B$ is subset of $A$ then $B$ is a subset of $A$ or $B= A$.

So in your second line you don't need to consider  $B= A$ as that is subordinate to $B$ being a subset of $A$.
So $B \subseteq A$ means:  For even $x\in B$, we have $x \in A$.  That's all.  There is NO point in consider the cases $B\subsetneq A$ and $B= A$ separately.  You could but there's no point.
.....
All said and done though, unfortunately there is no universal agreement about whether the symbole "$\subset$" means "unequal subset" or "subset (whether equal or not)".  I am firmly in the "subset  (whether equal or not)" camp.  In other words I beleive $B\subset A \iff B\subseteq A$.  But Im not the undisputed god arbiter of mathematical definition and notation.  There are those who will maintaing  $B \subset A \iff B\subsetneq A$.
I think one issue of confusion is that having four symbols:  $\subsetneq$, $\subset$, $\subseteq$ and $=$ is redundent.  If $subset$ means any subset, we have no need for the symbol $\subseteq$ as $\subseteq$ and $subset$ mean the same thing.  If $\subset$ means only non-equal subsets we have no need for $\subsetneq$ and $\subsetneq$ and $subset$ mean the same thing.
Perhaps the must clear and unambiguous method would be the use $\subsetneq, \subseteq$ and $=$ only.  That's what I would do if the $\subseteq$ symbol didn't require the cogneme of typing \subseteq instead of just \subset.  How/why should I be expected to always remember and emphasis that eq could be a possibility when 95% of the time it is not relevant?
Oh well.
Whatevs....
......
Then when you claim you aren't even sure what $A\cap B = B$ means.  It means just what it looks like it means  it means:   $A \cap B = \{x| x\in A$ and $x\in B\}$ is the same set of $B$.
I think maybe you are not getting that the result $A\cap B = B \iff B\subseteq A$ is supposed to be obvious.

If $A\cap B =B$ then even single element of $B$ is in the interesection of $A$ and $B$ and is in $B$.  That means $B$ is entirely contained in $A$.  So $B$ is a subset $A$.

That's all there is to it.  Formally right that up with element chasing.

If $A \cap B = B$ then if $x \in B$ then $x \in A\cap B$ so $x \in A$.  So for any $x \in B$ we have $x \in A$.  So $B\subseteq A$.  So $A\cap B = B\implies B\subseteq A$.


If $B\subset A$ then for any $x \in B$ then $x \in A$ and $x\in A\cap B$.  So $B = \{x|x\in B\} = \{x|x \in B$ and $x\in A\} = \{x|x\in A\cap B\} = A\cap B$.

That's all there is to it.
