The word contains is tricky, as it can mean both has as an element and is a superset of, depending on the context. Let's avoid it here.
Every set $A$ is a superset of $\emptyset$. But this is a theorem, not an axiom: there is a reason why this is true.
The definition of the subset relation is:
$B$ is a subset of $A$ (written $B\subseteq A$) whenever the following holds:
$$ \forall x \, \big(x\in B\Rightarrow x\in A \big)$$
Note that this is trivially satisfied when $B=\emptyset$, since $\forall x \, (x\notin \emptyset)$. So the empty set is a subset of any $A$ because all of the elements of the empty set are members of $A$.
Now, in this case, we have two sets: $A=\{\emptyset\}$ and $B=\{\{\emptyset\}\}$. Each of these sets has one element. In the case of $A$, that element is the empty set. In the case of $B$, that element is a set with a single element, that single element being $\emptyset$. Since the set whose only element is the empty set is the set $A$, it follows that $B=\{A\}$.
Now, when we put it like that, it is pretty clear that $A\in B$. But why is $A$ not a subset of $B$? Well, for $A$ to be a subset of $B$, it would need to be the case that every element of $A$ is an element of $B$. The set $A$ has a single element, the empty set. So $A$ is a subset of $B$ if and only if $\emptyset\in B$.
But as we have seen, $B$ has only one element, and that element is $A$, not $\emptyset$ (and we know that $A\neq \emptyset$ since $A$ is a set with one element and $\emptyset$ is a set with no elements).
So, since $\emptyset\in A$ but $\emptyset \notin B$, it follows that $A\not\subseteq B$.