A continuous curve intersects its 90 degrees rotated copy?

Some notation: For a curve (map) $\phi\colon[0,1]\to \Bbb C$, define the curve (set) $[\phi]:=\phi([a,b])$. Also, if $\phi(0)=0$, define $\phi^\pm$ as the concatenation of the given curve with its reversed negative: $$\begin{align}\phi^\pm\colon[a-b,b-a]&\to\Bbb C \\ t&\mapsto\begin{cases}-\phi(a-t)&t\le 0\\\phi(t+a)&t\ge0\end{cases}\end{align}$$

Now let $\gamma\colon[0,1]\to\Bbb C$ be our curve with $\gamma(1)=-\gamma(0)$. We can concatenate $\gamma$ with $-\gamma$ to a closed curve $\tilde\gamma$.

Fig 1. From $\gamma$ to $\tilde\gamma$.

Assume we can find a point $z_0\in [\tilde \gamma]\cap i[\tilde\gamma]$. Then $z_0$ is in one of $\pm[\gamma]$ as well as in one of $\pm i[\gamma]$. Therefore, one of the points $z_0, iz_0,-z_0,-iz_0$ is $\in [\gamma] \cap i[\gamma]$, as desired. Thus our goal is to find such an intersection $z_0$ of $[\tilde \gamma]\cap i[\tilde\gamma]$.

As the map $r\colon [0,1]\to \Bbb R$, $ t\mapsto |\gamma(t)|$ is continuous with compact domain, there exist $t_\min, t_\max\in[0,1]$ where it attains its minimum $r_\min$ and its maximum $r_\max$, respectively. If $i\gamma(t_\min)\in[\tilde\gamma]$, we can let $z_0=i\gamma(t_\min)$ and are done. Hence we assume from now on that $i\gamma(t_\min)\notin [\tilde\gamma]$. In particular, $r_\min >0$. Similarly, we may assume that $i\gamma(r_\max)\notin [\tilde\gamma]$.

Fix $r>r_\max$ and let $D$ be the open disk around $0$ of radius $r$. We say that a path $\eta\colon[a,b]\to\Bbb C$ escapes from $z$, if $\eta(a)=z$, $|\eta(b)|=r$, and $[\eta]\cap[\tilde\gamma]=\emptyset$.

Assume there exists a path $\eta\colon[0,1]\to\Bbb C$ that escapes from $0$. Because the complement of $[\tilde\gamma]$ is open, we can adjust $\eta$ locally to our liking without changing the end points or the disjointness to $[\tilde \gamma]$; therefore, we may assume that $\eta$ is a polyline (of finitely many segments). Now let $\zeta\colon[0,\ell]\to \Bbb C$ be a shortest path parametrized by length (so of length $\ell$) among all those with $\zeta(0)=0$, $|\zeta(\ell)|=r$, and $[\zeta]\subseteq [\eta^\pm]$. This is possible because $[\eta^\pm]$ is a simple graph with straight edges.

Fig 2. From a) general curve to b) polyline $\eta$, to c) shortest $\zeta$, to d) simple $\zeta^\pm$.

Then $\zeta^\pm$ is a point-symmetric path with end points on $\partial D$ and otherwise living in $D$. Also, $\zeta^\pm$ is simple: Any self-intersection comes from $t_1\ne t_2$ (wlog. $t_1<t_2$) with $\zeta(t_1)=\pm\zeta(t_2)$. But then the concatenation of $\zeta|_{[0,t_1]}$ and $\pm\zeta|_{[t_2,\ell]}$ would be shorter than $\zeta$, contradiction. The end points of $\zeta^\pm$ split $\partial D$ into two semicircle arcs. Together with either of these arcs, $\zeta^\pm$ forms a simple closed curve - a nice and friendly Jordan curve. Moreover, the interior regions of these two Jordan curves are disjoint, point symmetric to each other, and their union is $D\setminus [\zeta^\pm]$. By the point symmetry, $\gamma(0)$ and $\gamma(1)=-\gamma(0)$ are not in the same Jordan curve interior. It follows that $[\gamma]$ intersects $[\zeta^\pm]$, which is absurd.

We conclude that no path escapes from $0$.

Assume there is a path that escapes from $i\gamma(t_\min)$. Then together with the line segment from $0$ to $i\gamma(t_\min)$, we'd obtain a path that escapes from $0$, which we know does not exist. On the other hand, a line segment radially outward from $i\gamma(t_\max)$ escapes (here we use that $i\gamma(t_\max)\notin[\tilde\gamma]$). We conclude that any path from $i\gamma(t_\min)$ to $i\gamma(t_\max)$ intersects $\tilde \gamma$. In particular, this holds for the path between these points along $i\gamma$, thereby giving us an intersection point $z_0$, as desired.