Problem: Let $D\subseteq\mathbb C$ be open and $f:D\to\mathbb C$ continous such that $f$ is holomorphic on $D\backslash\mathbb R$. Show that $f$ is holomorphic on $D$ using Morera's theorem.
I have a question regarding this following proof: Let $\triangle\subset D$ be some triangle. Using Morera's therom we need to show that $$\oint_{\partial\triangle}f(z)dz=0.$$ The case $\triangle\cap\mathbb R=\emptyset$ is trivial so let's assume the intersection is non-empty. Then we cut the triangle into the following subsets: 
This way we get 3 sub-triangles that are completly contained in $D\backslash\mathbb R$ and some stripe $S$ of width $\epsilon>0$ around the real axis. For each of the triangles $t_{2,3,4}$ we obviously have $\oint_{\partial t_i}f(z)dz=0$. And since $\partial\triangle=\partial t_2\circ \partial S\circ\partial t_3\circ\partial t_4$ we just need to show that the integral along the border of $S$ yields $0$ when $\epsilon\to 0$ and that is my question: How do we conclude that? I would argue like this and want to know whether it is correct:
The shape of $S$ can be descriped as a trapezoid, let's denote the upper parallel edge with $\gamma_1$ and the lower one with $\gamma_2$. The non-parallel edges have a length that converges to $0$ as $\epsilon\to 0$. Also, as $\epsilon\to 0$, the edges $\gamma_1$ and $\gamma_2$ become somewhat inverse in lack of a better word (meaning that if I walk along either $\gamma_1\circ\gamma_2$ or $\gamma_2\circ\gamma_1$ I haven't moved at all). This should now yield $$\lim_{\epsilon \to 0}\left(\int_{\gamma_1}f(z)dz+\int_{\gamma_2}f(z)dz\right)=0$$ and therefore $\lim_{\epsilon\to 0}\int_{\partial S}f(z)dz=0$.