Int M is open and a manifold If M is an n-dimensional manifold with boundary, then Int M is an open subset of M , which is itself an n-dimensional manifold without boundary.
I am supposed to use these definitions: If M is an n-manifold with boundary, a point p in M is called an interior point of M if it is in the domain of an interior chart; and it is called a boundary point of M if it is in the domain of a boundary chart that takes p to $∂H^n$
How to do this?
I have shown that it is Hausdorff and second countable.
These are topological manifolds
We are not supposed to know that a point cannot be simultaneously a boundary point and an interior point.
 A: Here is my proof. Let me know if there is something that bother you.
Let M is a n-dimensional manifold with boundary. By definition, $Int M = \{p \in U \subset M | U \text{ open in } M \text{ and } \phi(U) \text{ is open in } \mathbb{R}^n \} $. So for every point on Int M, there always exist a open set U that is homeomorphic to open set in $\mathbb{R}^n$, therefore Int M is open in M. 
Because every openset of second countable space is second countable, and every openset of hausdorff space is Hausdorff, then Int M is second countable Hausdorff space. Therefore Int M is a n-dimensional manifold without boundary.
A: $Int M$ is defined as the set of all the points that belong to the domain of an interior chart. We want to see it is an open set in the topology of M. Take $p\in Int M$, let's see that there is a neighborhood of $p$ contained in $IntM$. 
We know, by definition, that there's a chart $(U,\phi)$ such that $p\in U\subseteq M$. But for any $q\in U$, $q$ must be an interior point, since it belongs to the domain of an interior chart: $(U, \phi)$. Therefore $q\in IntM$ and, since $q$ is arbitrary, $U\subseteq Int M$.
We saw that for any $p\in Int M$ there exists an open neighborhood $p\in U\subseteq IntM$, thus $IntM$ is open.
