Int M is open and a manifold

Question

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.

Answer 1

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.