Closure of simply connected set in $\mathbb{C}$ with smooth boundary is path-connected

Question

Let $\Omega \subset \mathbb{C}$ be a simply connected set with a smooth boundary (i.e., there is a differentiable function $\gamma : [a, b] \to \mathbb{C}$ so that $\gamma([a, b]) = \partial \Omega$). How can I show that $\overline{\Omega}$ is path-connected? We know that the boundary is path-connected via $\gamma$, and the interior is path-connected because an open, connected set is path-connected. I'm not sure how to connect these two parts via a path though. I did see this question, but it's a bit beyond my current knowledge.

Answer 1

It only remains the case of connecting a point $z_1 \in \Omega$ with a point $z_2 \in \partial \Omega$.

$\partial \Omega = \gamma([a, b])$ is compact because $\gamma$ is continuous, therefore $$ \{ t > 0 \mid z_1 + t \in \partial \Omega \} $$ has a minimum $t^*$. Then the segment $[z_1, z_1+t^*]$ connects $z_1$ with $z_3= z_1+t^* \in \partial \Omega$ in $\overline \Omega$. Now connect $z_3$ with $z_2$ with a restriction of $\gamma$.