How does the axiom of regularity make sense? From Wikipedia: 

The axiom of regularity is an axiom of Zermelo–Fraenkel set theory that states that every non-empty set A contains an element that is disjoint from A. 

I'm sure I'm grossly misunderstanding this, but it doesn't seem to make any sense. Two disjoint sets do not have elements in common, correct? So this says a set contains an element that it is disjoint with. So wouldn't this mean that the set doesn't contain that element if it is disjoint with it? To me, this axiom seems to be saying "Every non-empty set contains an element that it doesn't contain." This obviously doesn't make sense, so I'm definitely misunderstanding at least one aspect of this axiom, but which part?
 A: Your confusion seem to be due to levels of containment.
The members of a set is only those elements that are directly contained in the set. By this I mean that a set may contain other sets, but that does not make the elements of that set a member of the outer set.
This is a bit in contradiction in how we think of containment in ordinary life. Let's for example say you have a box of chocolates in your kitchen drawer. Then in ordinary life you would say that the drawer contains chocolates, but in mathematics terminology it doesn't it only contains a box (which in turn contains chocolates). Now you see that in mathematical terms the kitchen drawer and the box of chocolates are disjoint (but not in ordinary language).
To imagine a set that violates the axiom of regularity would probably need you to have some form of self containment. That is a container that in some way contains itself - which is quite contra intuitive, it doesn't fit well with the way we think of things.
A: As an example, let $X = \{1,2\}$ (where $0=\varnothing, 1= \{0\}, 2=\{0,1\}$). Then 
$$X\cap 1 = \varnothing$$
$$X\cap 2 = \{1\}$$
Due to the first equality, our set satisfies the axiom of regularity. $X$ being disjoint with $1$ does not imply that $X$ does not contain $1$; $X\cap 1 = \varnothing$ is a completely different statement from $1\notin X$. Indeed, the former holds, as  the only element of $1$ is $0$, which is not an element of $X$. 
A: In a nutshell, the axiom forbid "loops".
Consider he case $A= \{ a_0, a_1, a_2 \}$.
If the axiom does not hold, we have the possibility that: for all $i$ : $a_i \cap A \ne \emptyset$.
1) Consider $a_0$; if the intersection of $a_0$ and $A$ is not empty, it must be an element of $A$.
Thus, three possibilities: i) $a_0 \in a_0$: loop; ii) $a_1 \in a_0$; iii) $a_2 \in a_0$.
Two possibilities left to avoid loops.
2) Consider $a_1$; again, if the intersection of $a_1$ and $A$ is not empty, we have: i) $a_0 \in a_1$; ii) $a_1 \in a_1$: loop; iii) $a_2 \in a_1$.
But $a_0 \in a_1$ and the previous $a_1 \in a_0$ form a loop: $a_0 \in a_1 \in a_0$.
Thus what remain is: $a_2 \in a_0$ and $a_2 \in a_1$.
Now: 3) Consider $a_2$; we have: i) $a_0 \in a_2$; ii) $a_1 \in a_2$; iii) $a_2 \in a_2$: loop.
Again, with: $a_0 \in a_2$ and $a_2 \in a_0$ we have a loop and the same with 
$a_1 \in a_2$ and $a_2 \in a_1$.
