This is almost the same problem as in this question. However, the OP there was looking for a solution where we could assume any number of things, while I want to stick with just the given assumption (continuity). All the answers there addressed the OP's intent, making certain assumptions that might not be necessary. So none of them apply here.
Here's the question:
A continuous, injective curve $C$ goes from $(-1,-1)$ to $(1,1)$ in the Euclidean plane. Rotating $C$ in either direction by $90$ degrees with respect to the origin, we obtain a new curve $C'$ from $(-1,1)$ to $(1,-1)$. Prove that there exists a point $X$ that lies on both $C$ and $C'$.
In the linked question, the accepted answer by Arthur assumes every straight line through the origin intersects $C$ at exactly one point (of course, except the line $y=x$), and that $C$ goes strictly counterclockwise. Assuming these, he provides a proof. The other two answers are incomplete, each missing a case that they did not discuss.
Here's an approach that seems promising. Look at $C$ as a set of points, which we represent in polar coordinates $(r,\theta)$. There exists a point $P_0=(r_0,\theta_0)$ so that $r(P)\geq r_0$ for all $P\in C$ (by continuity), and similarly we have a upper bound with $P_1=(r_1,\theta_1)$. If we assume that there exists precisely two points in $C$ of distance $r$ to the origin for every $r$ so that $r_0<r<r_1$, then the difference of their $\theta$s is uniquely determined by $r$. Let's call this number $\theta(r)$. Since $\theta(\sqrt2)=\pi>\pi/2$ and $\theta(r_0)=0<\pi/2$, and $\theta$ is continuous, by IVT there is some $r$ so that $\theta(r)=\pi/2$. So we have a point on $C$ which is exactly $90$ degrees away from another point on $C$ of the same distance to the origin. Hence, rotating this point by $90$ degrees, the result still stays on the curve. We are done.
This approach still assumes more than just continuity. In particular, it is not in general true that there are exactly $2$ points of each $r$ on $C$. If we don't assume that, this proof breaks down since $r$ does not determine $\theta$. Any thoughts appreciated!
Best Answer
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 $\tilde\gamma$.
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.
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.