The boundary of a manifold is a closed subset. We want to show that the boundary $\partial M$ of an $n-$manifold M is a closed subset of the manifold. We show that its complement $M\setminus\partial M$ is open in $M$. Indeed,  each point $x \in M\setminus\partial M$ has an open neighborhood $V_x\subseteq M$ homeomorphic to $\mathbb R^n$. It remains only to show that $V_x$ lies entirely in $M\setminus \partial M$ which means that $V_x\cap \partial M=\emptyset$. Thank you for your help! 
EDIT: In a manifold with boundary each point has an open neighborhood that is homeomorphic to $\mathbb R^n$ or to $\mathbb R^n_+=\{(x_1,\cdots,x_n)\in\mathbb R^n\;|\; x_n\ge 0\}$, the points who have open neighborhoods homeomorphic to $\mathbb R^n_+$ form the boundary of the manifold.

This is obvious. Let $x \in M \setminus \partial M$. There exists an open neighborhood $V$ of $x$ in $M$ which is homeomorphic to $\mathbb R^n$. Then any $y \in V$ has the same property, hence $V \subset M \setminus \partial M$.
By the way, your definition of the boundary is not correct. Take $M = \mathbb R^n_+$. Then each $x \in \mathbb R^n_+$ has $\mathbb R^n_+$ as an open neighborhood which is homeomorphic $\mathbb R^n_+$. A correct definition is this: A point $p \in M$ is called a boundary point if there
exists a homeomorphism $\phi : U \to \mathbb R^n_+$ defined on an open neighborhood $U$ of $p$ in $M$ such that $\phi(p) \in \partial \mathbb R^n_+ =\mathbb R^{n-1} \times \{ 0\}$. 
