Is it The Axiom of Power Set that guarantees the existence of (all) subsets? The Axiom of Power Set asserts that:

For any set $S$, there exists a set $\wp$ such that $X \in \wp$ if and only if $X\subseteq S$.

That is, if something is a subset of $S$, then it's a member of $\wp$, and $\wp$ contains nothing but subsets of $S$. 
However, the way I see it, the axiom does not guarantees the existence of subsets of $S$ at first. Possibly we can tell whether something is a subset of $S$ or not, but how can we ascertain the existence of all of them?
 A: I believe this question can be answered by  another thorough read of the axiom. Note that it says, 

for every $S$, there exists a set $P(S)$ such that ...etc.

The whole point of the axiom is to declare the existence of the set of subsets. I'll elaborate:  
We have the definition: $X$ is a subset of $S$ (notation $X\subset S$) $\iff$ $(\forall x)(x\in X\implies x\in S)$.
Since sets exist, this is a solid definition. Using this definition, we can see that subsets (of any set) exist, because $\emptyset $ and $S$ always satisfy this definition. How many there are in total, we don't know yet. 
When we make the axioms, we are deciding which objects we will call sets, and which objects we will not call sets. At a certain point we decide that we don't want $\{ X\mid X\subset S\}$ to be a proper class (i.e. not a set).  So we create the rule (axiom) that $\{ X\mid X\subset S\}$ is a set, for all sets $S$. For convenience we use the notation: $P(S)=\{ X\mid X\subset S\}$.
Now let's sum this up:


*

*Subsets exits, because sets exist.

*Any subset of a set $S$, will belong to $P(S)$, because that's simply how we defined $P(S)$.

*The Power Set Axiom now says: Henceforth, we will all agree that we consider $P(S)$ to be a set for every set $S$. 

