Q:

Why is $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ true?

My book "Introduction to SetTheory" says 

$\{\emptyset\}\in\{\{\emptyset\}\}$ but $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$

When we say $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$, we mean that $\emptyset$ is not listed amongst the elements of $\{\{\emptyset\}\}$.
I was under the impression that every set contains the empty set. Hence, the elements of $\{\{\emptyset\}\}$ would be $\emptyset,\{\emptyset\}$. If that were true, wouldn't the assertion that $\{\emptyset\}\not\subseteq\{\{\emptyset\}\}$ be false?

A:

$\varnothing\ne\{\varnothing\}$. The second is the only element of $\{\{\varnothing\}\}$. Do you see $\varnothing$ by itself listed in there? No.
When we say every set contains the empty set, we mean as a subset, not as a member or element, so we are saying that $\varnothing\subseteq X$ not $\varnothing\in X$. Do you think $\varnothing\in\varnothing$ for instance? That's not possible!

A:

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 $A\not\subseteq B$.

