This is a similar question to being asked to prove that $\emptyset \subseteq A$ for any set $A$. To prove it, you need to show if $x \in \emptyset$, then $x \in A$. But the statement $p \implies q$ is true even when the statement $p$ is false, because $False \implies True$ is a true statement. And for any $x$, $x \in \emptyset$ is a false statement. (Note that $False \implies False$ is also a true statement, so it does not matter if $x \in A$.)
Having said that, to prove $A \iff B$, you need to prove $A \implies B$ and $A \impliedby B$.
Let's prove the $\implies$ direction first: suppose $A \subseteq \emptyset$. We want to prove $A = \emptyset$. Suppose by contradiction that $A \neq \emptyset$. Since $\emptyset \subseteq A$, then there must exist $x \in A$ such that $x \not \in \emptyset$. But if $x \in A$ and $x \not \in \emptyset$, then $A \not \subseteq \emptyset$, which contradicts our assumption.
Now for the $\impliedby$ direction: suppose $A = \emptyset$. Then since any set is a subset of itself, we have that $A \subseteq \emptyset$, as desired.