For any closed smooth curve $z:[a,b]\to \Bbb C$, we define the interior $\operatorname{Int}(z)$ of $z$ as follows: Because $\operatorname{Im}(z)$ is bounded, we can find a circle $C$ such that $\operatorname{Im}(z)$ is contained in the interior of $C$, fix a point $v_0$ outside the circle $C$, $\forall w\in \Bbb C-\operatorname{Im}(z)$, if there exists a path (a continuous map $[0,1]\to \Bbb C$) from $w$ to $v_0$ which doesn't intersect $\operatorname{Im}(z)$, we say $w\notin \operatorname{Int}(z)$, otherwise $w\in \operatorname{Int}(z)$.
$1.$ How to show $\operatorname{Int}(z)$ is not empty?
$2.$ How to show $\operatorname{Int}(z)$ is open?
For 2, notice that $\mathrm{Im}(z)$ is compact. So, given $w\not\in\mathrm{Im}(z)$, there is a ball centered at $w$ that does not intersect $\mathrm{Im}(z)$. The points in this ball can be connected to $w$, so either they all belong to $\mathrm{Int}(z)$, or none of them does.