The Scwarz Reflection Principle says that if $A$ is an open, connected set in the upper half plane of $\mathbb{C}$ whose boundary intersects the real axis in an interval $[a,b]$, then we can extend a holomorphic function defined on $A$ (and continuous on $A \cup (a,b)$) to a holomorphic function defined on $A \cup (a,b) \cup A^*$, where $A^*$ is the reflection of $A$ over the real axis.
I'm not confused about the analysis aspect of this theorem, but there's one topological wrinkle I want to iron out. How do we know that $A \cup (a,b) \cup A^*$ is open? It's an intuitive fact, but I can't figure out how to prove it rigorously.
My idea is to pick a point $x \in (a,b)$ and show that there exists $\epsilon > 0$ such that the set $\{z : |z-x| < \epsilon, \operatorname{Im}(z) > 0\}$ is contained in $A$; I'm not sure where to go from here.
You cannot prove that, since it is false. Consider the set$$A=\{z\in\Bbb Z\mid\operatorname{Im}(z)>0\}\setminus\left\{\frac in\,\middle|\,n\in\Bbb N\right\}.$$Then $A$ is an open set and $\partial A\supset\Bbb R$. But $A\cup\Bbb R\cup A^*$ is not an open set. More generally, if $a<0<b$, then $A\cup(a,b)\cup A^*$ is not an open set. For instance, it contains no neighborhood of $0$.
I am not claiming that there is something wrong with the Schwarz Reflection Principle. But it is usually assumed from the start that $A\cup(a,b)\cup A^*$ is an open set.