How does the axiom of regularity forbid self containing sets?

The axiom of regularity basically says that a set must be disjoint from at least one element. I have heard this disproves self containing sets. I see how it could prevent $A=\{A\}$, but it would seem to do nothing about $B=\{B,\emptyset\}$. It is disjoint from $\emptyset$. What is a disproof of the existence of $B$, and how is it related to the axiom of regularity?

Let $A$ be any set. Then $\{A\}$ is a set, and by regularity $\{A\}$ must contain an element disjoint from $\{A\}$. The only element of $\{A\}$ is $A$, so $A\cap\{A\}=\varnothing$, and it follows immediately that $A\notin A$.

No. The axiom of regularity says this: every non-empty set contains an element disjoint from it. So suppose there were a set $B$ such that $B = \{ B, \emptyset \}$; then the set $\{ B \}$ contains no element disjoint from $\{ B \}$: $$B \cap \{ B \} = \{ B \} \ne \emptyset$$ so there is no such set $B$.

The axiom of foundation or regularity alone can not show that there is no $x$ such that $x \in x$.

The axiom of regularity (also called the axiom of foundation) asserts that every set has a $\in$ minimal element. That is, for all $x$, there exists a $y \in x$ such that there are no $z \in x$ with $z \in y$.

The axiom of foundation is not exactly equivalent to the fact that there does not exists an $x$ such that $x \notin x$. There is a two element model of foundation (extensional, union, and pairing) such that there exists a $x$ with $x \in x$. Let $M = \{x,y\}$ where $x$ and $y$ two different objects. Define $\in^\mathcal{M} = \{(x,y), (y,y)\}$. You can verify foundation holds in this model $\mathcal{M}$, but $y \in y$.

Note that the axiom of comprehension and foundation can show that there is no $x$ such that $x \in x$

Together with axiomschema of separation:

Assume that $A\in A$, then there's a set $B = \{x\in A: x=A\}$ which is non-empty since $A\in B$. Now for any $x\in B$ it's true that $x=A$, so $\forall x\in B: x\cap B = A\cap B$, but since $A\in A$ and $A \in B$ it would be that $A\cap B \ne \emptyset$. But that's equivalent to $\neg\exists x\in B: x\cap B = \emptyset$ which contradicts the axiom of regularity.

Therefore we can conclude $A\notin A$.

Here is a formal proof in Fitch: