Why is $\emptyset$ a subset of every set? Please read the details of this question before answering.
I am reading a book in which they say $\emptyset \subseteq S$ for any set $S$.  One justification they give for this is that if $\emptyset \not \subseteq S$, then there must be some element in $\emptyset$ that is not in $S$.  And we conclude that this cannot happen since $\emptyset$ has no elements.
.....But to prove $\emptyset \subseteq S$, we need to show every element of $\emptyset$ is in $S$.  But this is impossible too, since $\emptyset$ has no elements.
I don't get it.  By the way, I am familiar with vacuous truth (i.e., that $\mathrm{false} \implies \mathrm{true}$ is a true statement).  But if we reason the above statements in the way I did, then something isn't right.  The question is, what isn't right?
 A: If $\varnothing$ has no elements, then we don't have to do any work to show that all of them are elements of $S$!
Anyways, you say that we need to show
$$ \forall x: x \in \varnothing \implies x \in E$$
but you say you can't do this, because $x \in \varnothing$ is identically false. Can you articulate why you think that's a valid argument?
A: The difference between $\emptyset \subseteq S$ and $\emptyset \nsubseteq S$ is that the former states that "every element in $\emptyset$ is also an element in $S$" while the latter states that "some element $\emptyset$ is not in $S$". 
The second statement is clearly false, since if it were true then you should be able to exhibit some element from the empty set which is not in $S$. Of course, you can't do that since there are no elements in the empty set. 
The first statement is true, since it were false, then you should be able to exhibit some element from the empty set which is not in $S$. Of course, you can't do that since there are no elements in the empty set. 
So, I guess your confusion stems from wrongly appealing to 'vacuously true' without paying enough attention to the details. A correct application of vacuous implication would be to conclude that by the same argument it holds that $\emptyset \subseteq S^c$, where $S^c$ is the complement of $S$ relative to some universe. This is of course correct, and essentially the same argument. 
I hope this helps. 
A: The rules of logic say a false statement implies anything. Hence $x\in \emptyset\implies x\in E$ for any set $E$.
