How to show that the winding number must be $0$ or $\pm$ $1$ in this case? Prerequisite:
$K: [0,1] → \mathbb{C}$ piecewise, continuously differentiable, closed path that meets the non-positive real axis at only at one point
$K^{-1} (\{x|x \le 0\})= \{ t_0 \}$ with $0 < t_0 < 1$ and $K(t_0)=−r < 0$ and does not meet at the origin.
Show:
$\eta (0) \in \{0, \pm 1 \}$ (where $\eta$ is the winding number),
and $0, 1$, and $− 1$ occur if $K$ runs locally at $t_0$ on one side of the negative real axis or intersects the negative real axis from top to bottom or rather from bottom to top.
I tried to prove by contradiction, that if we assume, that $n=2$, then the closed path would obviously intersect the negative real axis twice at two real numbers $w=(a,0)$ and $z=(b,0)$. Would that be enough to prove the statement?

Pick a continuous branch $\theta(t)$ of $\operatorname{Arg}\gamma(t)$ such that $\theta(t_0)=\pi$. By assumption, $\theta(0),\theta(1)\notin\{\ldots,-3\pi,-\pi,\pi,3\pi,\ldots\}$. By intermediate value theorem, the values of $\theta(0)$ and $\theta(1)$ must lie inside the open interval $(-\pi,3\pi)$, otherwise $\theta(t)$ will be equal to $-\pi$ or $3\pi$ for some $t\in[0,1]$. Hence $|\theta(1)-\theta(0)|<4\pi$ and it must be equal to $0$ or $2\pi$. This means the winding number of $\gamma$ about the origin is $0$ or $\pm1$.
