Intersection of Set A and a set containing empty set

In general,

$$\emptyset \cup S = S \not=\{\emptyset\} \cup S$$

Note that in our case, $A = \{\emptyset, \{\emptyset\} , \{\{\emptyset\}\}\}$, $A $ is a set of sets and $\emptyset $ is just a set with nothing. $B = \{\emptyset\}$ is also a set of sets, but happens to only contain the empty set. Then we have

$$A = \{\emptyset, B, \{B\}\} $$

And thus $A \cap B = B = \{\emptyset\}$ given that $B \subset A $.

Also, for the $A $ given, $A \cup B = A $ only because $B \subset A$. There are plenty of sets $S $ where $S \cup B \not= S$.

"Nothing exists" is one (tongue in cheek) way of thinking about the axiom of the empty set.

The empty set $\emptyset$ exists by fiat; which is to say (either directly as an axiom or derived from other axioms), $\exists \emptyset : \forall x, x \notin \emptyset$.

Once we have any set $x$, we have rules by which we can prove the existence of other sets; such as the set $\{x\}$ whose only element is $x$, and so on. $\emptyset$ is no different in this respect; it is a set, and we have rules that show how to construct other sets from sets whose existence we have already proven.

The union of two sets $A$ and $B$ is a set with the property $\forall x: (x \in A \cup B \iff x \in A \lor x \in B)$. So thinking about $\emptyset \cup A$, we are talking about a set:

$$\forall x: (x \in \emptyset \cup A \iff x \in \emptyset \lor x \in A)$$

But by definition, it can never be true that $x \in \emptyset$; so the above boils down to:

$$\forall x: (x \in \emptyset \cup A \iff x \in A)$$

or in other words (see axiom of extensionality); $\emptyset \cup A = A$.

On the other hand there does exist some set $x$ s.t. $x \in \{\emptyset\}$; namely, the set $x = \emptyset$. So when we consider $\{\emptyset\} \cup A$,we are talking about the set:

$$\forall x: (x \in \{\emptyset\} \cup A \iff x \in \{\emptyset\} \lor x \in A)$$

Thus $\{\emptyset\} \cup A$ is not necessarily equal to $A$; because (depending on $A$) we might not have $\emptyset \in A$, but we do have $\emptyset \in \{\emptyset\}$; and therefore $\emptyset \in \{\emptyset\} \cup A$.

Note that $\emptyset=${ } which contains no element while the set $X=${$\emptyset$} is non-empty containing an element $\emptyset$.

{$\emptyset$}$\cap A$ represents the intersection of $X$ with $A$ which is clearly the set $X$ itself as $X\subset A$