Peano Induction Axiom This is a typical rendition the Peano Axiom of Induction:

If subset $S \subseteq \mathbb{N}$ contains $1$ and is closed under the successor function (i.e., $n \in S$ implies $\sigma\text{n} \in S$ where $\sigma$ is the successor function) then that subset $S$ is all of $\mathbb{N}$.

How do we make the leap that $S$, which might not be all of $\mathbb{N}$ eventually does cover all of $\mathbb{N}$? I can only assume $\sigma\text{n}$ takes things outside of $S$?
How do we make the leap that $S$, which might not be all of $\mathbb{N}$ eventually does cover all of $\mathbb{N}$? I can only assume $\sigma n$ takes things outside of $S$?


When you say that "$S$..might not be all of $\mathbb{N}$", you are actually trying to find out whether $S$ and $\mathbb{N}$ are identical or not. In other words, you need a proof of the question whether $S=\mathbb{N}$ or not. But you can only prove the equality between two sets when you know that definition of each set. So, you need to answer first,


  What is the definition of $\mathbb{N}$?
The Peano Induction Axiom can be thought of as one of the 5 essential properties of the natural numbers. See Peano's Axioms. (There are other, less widely used formulations.)

The Induction Axiom says that, if you have a subset $P$ of $\mathbb{N}$, then we can show that all elements of $\mathbb{N}$ will be in $P$ if

(a) $1 \in P$, and 

(b) if $x \in P$, then the successor of $x$ will also be in $P$