How is an empty set truly "empty"? In a related question, an answerer says:

an empty bag is a bag with nothing inside it.

Makes sense, but I'm reading a textbook right now that says:

The empty set has only one subset (namely, itself).

Which seems odd to me. If a set is empty, why would it contain a subset? Regardless of whether or not the subset is also an empty set, the "empty set" still contains a subset, so how is it really "empty"?
To go back to the empty bag analogy, it just sounds like my text book is saying:

an empty bag is a bag with nothing but an empty bag inside it.

Which doesn't make sense because then the bag isn't empty. I can't wrap my head around this concept.
 A: Your last statement "an empty bag is a bag with nothing but an empty bag inside it." is not correct.
Because, A is a subset of B does not mean that B contains the set A. It simply means that A is another bag(container) like B(which is also a container) which contains the items taken from inside the bag B.
A: The use of the word "contains" is a bit misleading, there. When we talk about a set $A$ "containing" a subset $B$, what we really mean is that $A$ contains all elements of the set $B$ (that is, for every $x\in B,$ we have $x\in A$). In that sense, we're saying that an empty bag is a bag that contains exactly what an empty bag contains: nothing at all.
It is possible (though not in the case of the empty set) that a set $A$ contains a set $B$ both as an element and as a subset. For example, the set $$A=\bigl\{1,\{1\}\bigl\}$$ has $$B=\{1\}$$ as an element (obvious, hopefully) and as a subset (because $A$ contains every element of $B$). Here's (one place) where the grocery bag analogy breaks down, however, since the grocery bag analogy would suggest that the above $B$ is a subset of the above $A$ by virtue of being contained in $A$ as an element. This is not so. Indeed, if we consider $$C=\bigl\{\{1\}\bigl\},$$ then we find that $B$ is an element of $C,$ but not a subset of $C,$ since $1$ is not an element of $C$!    
So, given two arbitrary sets $A$ and $B$, $B$ may be:


*

*a subset of $A$ but not an element (e.g.: for any set $A$ such that $\emptyset\notin A$, let $B=\emptyset$),

*an element of $A$ but not a subset,

*an element and a subset of $A$,

*neither an element nor a subset of $A$ (consider $A=\{1\},$ $B=\{3\}$).

A: Do you know what it really means for one set to be a subset of another set? If the set $A$ is a subset of the set $B$, then we write $A\subseteq B$; now, to show that $A\subseteq B$, we must show that $x\in A\to x\in B$. 
The fact that $\varnothing\subseteq\varnothing$ is really not that surprising if you are aware of what a so-called "vacuous truth" is. That is, $x\in\varnothing$ is clearly not true; thus, we can freely conclude (i.e., "vacuously") whatever we want. Hence, $x\in\varnothing\to x\in A$, where $A$ is any set whatever, including the empty set.
Does that make things any clearer? It seems like you want a materialistic or intuitive way of seeing that $\varnothing\subseteq\varnothing$ when really the key lies in understanding what it means for one set to be the subset of another set and also what a vacuous truth is and how you can use that. 
A: You are confusing "subset of" ( $\subseteq$ ) with "element in" ( $\in$ ).
This is possibly because you are confusing the set of elements (marbles) with the container (the bag) which holds the set.   The bag is not the set, the set is the collection held by the bag.
The set is empty, as there are no marbles held by the bag.   This empty set cannot be subdivided into smaller sets; as none is as few marbles as you can get.
Thus the empty set is the only subset of the empty set, $\varnothing \subseteq \varnothing$, but the empty set is not an element of itself. $\varnothing \notin\varnothing$.
Conversely, the set of an empty set is not an empty set. $\{\{\}\}\neq \{\}$
A: As other posters have noted you are confusing "element of" with "subset of".
To apply your bag/marble analogy to this situation, if we say that A is a subset of a set B then we are saying the following things:


*

*A and B are the same kind of thing. That is, both A and B are bags and neither of them are marbles.

*x ∈ A implies x ∈ B. That is, all of the marbles in bag A are also to be found in bag B. For example, perhaps A contains a red marble and a blue marble, while B contains a red marble, a blue marble, and a green marble.

A: In my opinion, OP's confusion arises from the fact that the bag analogy suggests that if bag $B$ is in bag $A$, then bag $B$ is a subset of $A$ (because all things in bag $B$ are, physically, in bag $A$ as well). This is where the bag analogy breaks down. The fact that a set $B$ is in a set $A$ does NOT imply that $B$ is a subset of $A$. For example, let $B=\{2\}$ and $A=\{B, 1\}$. Then $B\in A$ but $B\nsubseteq A$. The reason why $B\nsubseteq A$ is because there's at least one thing that is in B but is NOT in A. One such thing is $2$ (in this case, it's the only one). $A$ contains only $B$ and $1$. The fact that $B$ itself contains $2$ doesn't, in Mathematics at least, imply that $2$ is in $A$. $A$, again, contains only $B$ and $1$, not $2$.
