I came up with the following and wanted to be sure if my proof is correct:
Theorem: Let $G$ be an open set in $\mathbb{C}$ and $a \in G$ with $B_X(a,r)\cap \delta G = \emptyset$. ($\delta G$ is the boundary of $G$). Then $B_X(a,r) \subseteq G$.
Proof: Suppose not. Then we can pick $z \in B_X(a,r)$ with $z \notin G$. The line segment $[a,z]$ is contained in $B_X(a,r)$, since balls are convex.
Put $f(t)= a(1-t) + zt, t \in [0,1]$
Put $t_0 := \inf\{t : f(t) \notin G\}$, which exists because $f(1) = z \notin G$ and note that $f(t_0) \notin G$ by the continuity of $f$. Also note that $t_0 \neq 0$, since $a=f(0) \in G$. Thus, we can take a sequence $0 \leq t_n, n \geq 0$ such that $t_n \nearrow t_0$ (strictly) and by the continuity of $f$ we have $f(t_n) \to f(t_0)$. However, $f(t_n) \in G$ for all $n$ and hence $f(t_0)$ is in the closure of $G$. But this means that $f(t_0) \in \overline{G} \cap G^c = \overline{G} \cap \overline{G^c}= \delta G$, because $G$ is open. This contradicts the hypothesis $B_X(a,r) \cap \delta G=\emptyset$.
Is this correct?
If $B_X(a,r) \cap \partial G= \emptyset$, $B_X(a,r)$ cannot intersect $G^\complement$ or the ball would be disconnected (by the disjoint cover $G$ and $\Bbb C\setminus \overline{G}$, intersected by $B_X(a,r)$.
This argument works for any connected neighbourhood of $a$.