First let us see what does that mean $X\iff Y$. The meaning is:
if $X$ is known to be true then $Y$ can be shown true, and if $Y$ is known to be true then $X$ can be shown true as well.
The above can be plainly written as $X\implies Y\land Y\implies X$ where $\land$ is "logical and" (i.e. conjuction) connective.
In the question above we have $X\iff Y\iff Z$, which translates to: $X\iff Y \land Y\iff Z$
From this we can prove that $X\iff Z$ using $Y$ as an interpolant. That is assume that $X$ is true, prove $Y$ is also true, and use the proof of $Y\iff Z$ to show that $Z$ is true, and similarly for the other direction.
We need, if so, to prove three equivalences, each by proving a double implication. We can reduce this into two equivalences:
- $X\iff Y$,
- $Y\iff Z$.
As we have seen how this would prove $X\iff Z$ as well. This means we need only two equivalences instead of six, which translates as four implications. However we can instead prove these three:
- $X$ implies $Y$,
- $Y$ implies $Z$,
- $Z$ implies $X$.
From this we can deduce that $Y\implies X$, this time by using $Z$ as an interpolant, and the same for $Z\implies Y$. (It can be shown that no less than three proofs are sufficient, this is a graphical proof really: you cannot close a path between three points with only two edges.)
To prove an implication $X\implies Y$ we assume that $X$ is true, and show that in this case $Y$ is also true.
From here things should be clearer, and we can actually approach the proof.
Assume that $A\subseteq B$, in this case $A\cap B=\{a\in A\mid a\in B\}$ however our assumption was that $a\in A\rightarrow a\in B$, therefore $A\cap B=A$.
Assume $A\cap B=A$, now $a\in A\cup B$, either $a\in A$ or $a\in B$. If $a\in A$ then $a\in A\cap B$ therefore $a\in B$. We conclude that $a\in A\cup B\rightarrow A\in B$. From this $A\cup B=B$ follows.
Assume $A\cup B=B$. Take $a\in A$ therefore $a\in A\cup B$, from our assumption $a\in B$. By definition we have that $A\subseteq B$.
In each of the three steps, we assumed a condition is being met and proved something else to be true as well. From this we can see how assuming $A\subseteq B$ will give us a proof that $A\cup B=B$. We simply "combine" the proofs.
Therefore to show that $A\cap B=A\implies A\subseteq B$ we can use the proofs in (2) and (3) combined, and similarly to have that $A\cup B=B\implies A\cap B=A$.