This is problem 8-3 from John Lee's Introduction to Topological Manifolds. Invariance of the boundary, 2-dimensional case: Suppose $M$ is a $2$-dimensional manifold with boundary. Show that a point of $M$ cannot be both a boundary point and an interior point.
I try to use the following theorem from problem 8-1 : Suppose $U \subset \mathbb{R}^2$ is an open subset and $x \in U$. Then $U - \{x\}$ is not simply connected.
Suppose $p$ is both an interior and boundary point. Choose coordinate charts $(U,\phi)$ and $(V,\psi)$, $\phi(U)$ is open in Int$\mathbb{H}^2$ and $\psi(V)$ is open in $\mathbb{H}^2$, with $\psi(p) \in \partial \mathbb{H}^2$. Let $W=U \cap V$; then $\phi(W)$ is homeomorphic to $\psi(V)$. From the above problem, $\phi(W) - \phi(p)$ is not simply connected. So we would reach a contradiction if $\psi(V) - \psi(p)$ is simply connected. However, I do not know if this is true. How could I show this problem?
I will keep your notation: Withot loss of generality we can assume that $\psi(p)=0$. We have that $0\in \psi(W)$, so we can choose $r>0$ such that $$ B(0,r)\cap \mathbb{H}^2 \subseteq \psi(W), $$ where $B(0,r)$ is the open ball in $\mathbb{R}^2$ centered at the origin with radius $r$. Let $U'=\psi^{-1}(B(0,r)\cap \mathbb{H}^2)$, thus $\phi(U')\setminus\{\phi(p)\}$ is not simply connected (by the theorem you mentioned). We have to prove that $U'\setminus\{p\}$ is simply connected, in contradiction with $\phi$ being a homeomorphism. But note that $U'\setminus\{p\}$ is homeomorphic to $B(0,r)\cap \mathbb{H}^2\setminus\{0\}$, so I will prove that $B(0,r)\cap \mathbb{H}^2\setminus\{0\}$ is simply connected.
Let $x_0\in B(0,r)\cap \mbox{int}\mathbb{H}^2$ and let $B(0,r)\cap \mathbb{H}^2\setminus\{0\}=:X$, and define $F:X\times [0,1]\to X$ by $$ F(x,t) = x_0 + (1-t)(x-x_0), $$ it is easy to see that $F$ is a strong deformation retraction of $X$ onto $\{x_0\}$, and thus $X$ is simply connected.