Proof of No Recursion implied by Axiom of Regularity

And it claims the Axiom of regularity implies that no set can be a member of itself. I don't follow whats being conveyed here. Consider this example:

Let $L = \lbrace{1 , L} \rbrace$

or to do this in ZFC

Let $L = \lbrace \lbrace \emptyset \rbrace , L \rbrace$

Then consider $L \cap \lbrace \emptyset \rbrace$

Obviously this is empty. So we meet the disjointedness condition, yet we can still have self membership. What am I missing here?

The Axiom of Regularity states that if $A$ is a non-empty set, then there is $B\in A$ with $A\cap B=\emptyset$.
In your example, consider $A=\{L\}$. Is there $B\in A$ such that $A\cap B$ is empty.