I'm new to manifolds and trying to get a handle on the concepts.
Suppose I have a subset S of $\mathbb R^n$. I want to see if S is a n-manifold of $\mathbb R^n$.
So if S is a n-manifold, then every point of S has a neighborhood that is topologically equivalent to an open unit ball in $\mathbb R^n$.
So topologically, isn't this equivalent to saying every point of S is an interior point of S (ie: S has no boundary points)?
So am I right in thinking that a subset S of $\mathbb R^n$, is an n-manifold if and only if every point of S is an interior point?
You are right, $S$ must be an open subset of $\mathbb R^n$. However, this is not at all trivial.
What you need to prove it is invariance of domain. This is a theorem which says that if $U$ is an open subset of $\mathbb R^n$ and $f : U → \mathbb R^n$ is an injective continuous map, then $f(U)$ is open and $f$ is a homeomorphism between $U$ and $V$. See https://en.wikipedia.org/wiki/Invariance_of_domain.
In the present case we know that each $x \in S$ has an open neighborhood $V$ in the space $S$ which is homeomorphic to an open subset $U \subset \mathbb R^n$ (you can even require that $U$ is the open unit ball), but a priori there is no reason why $V$ must be open in $\mathbb R^n$. That is the point where we must invoke invariance of domain.
Note that for smooth manifolds the argument is simpler. In fact, $V$ is diffeomorphic to $U$ whence we can conclude that $V$ is open in $\mathbb R^n$. See any textbook on multivariable calculus.