The interior of open set in a convex set is not empty

$\newcommand{gae}[1]{\newcommand{#1}{\operatorname{#1}}}\gae{cl}\gae{int}$ The topological property you are looking for is that $H^n$ is contained in the closure of its interior. In fact, this property is necessary and sufficient for your condition to hold. Let $X$ be a topological space and $S\subseteq X$. If $\cl_X\int_X S\supseteq S$, then every open set $U$ such that $U\cap S\ne\emptyset$ satisfies $U\cap\int_X S\ne\emptyset$ as well. Since $U\cap \int_X S$ is open in $X$ and $U\cap \int_X S\subseteq U\cap S$, we have that $U\cap \int_X S\subseteq \int_X(U\cap S)$. Vice versa, suppose that $S\setminus \cl_X \int_X S\ne \emptyset$. Then, $S\setminus\cl_X\int_X S$ is a non-empty subset of $S$ which is open in $S$ and which cannot contain open subsets of $X$ (because $(S\setminus \cl_X\int_X S)\cap\int_X S=\emptyset$). Therefore $\int_X(S\setminus\cl_X\int_X S)=\emptyset$.

By definition of the relative topology, if $U$ is open in $H^n$ it exists an open subset $V$ of $\mathbb R^n$ such that $U = V \cap H^n$. If $U$ is not empty, it exists $x \in U$. Therefore $x \in V$ and it exists an open ball $B(x,r)$ centered on $x$ with $B(x,r) \subseteq V$.

If $x=(x_1, \dots, x_{n-1}, 0)$ then $B(\bar x, r/4) \subseteq B(x,r) \subseteq U $ where $\bar x = (x_1,\dots,x_{n-1}, r/2)$. And if $x=(x_1, \dots, x_{n-1}, x_n)$ with $x_n >0$ then $B(x, \bar r) \subseteq B(x,r) \subseteq U $ where $\bar r = \min(r, x_n/2)$. Proving that the interior of $U$ is not empty.

Here, the main argument is that for an open ball $B(x,r) \subseteq \mathbb R^n$ we have $B(x,r) \cap H^n = B(x,r)$ for $r$ small enough and $x_n >0$.

This is not related to the fact that $H^n$ is convex. For example $L= \{(x,0) \mid x \in \mathbb R\}$ is a convex subset of the plane $\mathbb R^2$. $I= \{(x,0) \mid x \in (0,1)\}$ is an open subset of $L$. However, the interior of $I$ is empty in $\mathbb R^2$.