Will a path between $(x, y)$ and $(-x, -y)$ always intersect a 90 degree rotated copy?
This proof assumes there is no "backtracking" or radial movement (i.e. every line through the origin intersects the curve in exactly one point, with the obvious exception of the line through $(x, y)$ and $(-x, -y)$, which intersects the curve twice). Also, the curve is continuous and goes counterclockwise around the origin from $(x, y)$ to $(-x, -y)$.
Take the line through the origin and $(-y, x)$ ($90^\circ$ counterclockwise rotated copy of $(x, y)$). If the curve intersects the line at $(-y, x)$ we are done. If not, it intersects the line either outside or inside $(-y, x)$. Let's say inside (like your yellow, green and orange examples). Then the $90^\circ$ rotated curve goes on the inside of $(-x, -y)$.
This means that if we take a continuous "sweep" of lines through the origin, starting with the one through $(-y, x)$ and ending with the one through $(-x, -y)$, then the original curve goes from being on the inside of the rotated curve to being on the outside. By the intermediate value theorem they must intersect.
Edit $\quad$ This answer misses the situations where a curve does not fall under the two cases mentioned (see the following image for two examples). It is also possible that $Y$ from the first case is empty (for instance, when $D=C$). However, this does not affect the outcome. One last point; I realize that the third assumption is not so easily justified (take as an example a curve that intersects itself infinitely many times). We, therefore, need a specific definition of a curve that does not allow infinite self-intersections to avoid complications.
Here is an idea:
We can make the following assumptions:
A.1 $\quad$ As YiFan says, we can assume without loss of generality that the points are $A=(-1,0)$, $B=(1,0)$, $A'=(0,-1)$, and $B'=(0,1)$.
A.2 $\quad$ The curve $C$ connecting $A$ to $B$ does not pass through $A'$ or $B'$. (The statement is trivially true in such cases.)
A.3 $\quad$ The curve $C$ does not intersect itself. (Given such a curve, we can simply take another curve that is a subset of it with no self-intersections.)
We then have two cases:
Case 1 $\quad$ The curve $C$ passes "below" $B'$ and "above" $A'$. Let $D$ be $C$ rotated about the origin $180$ degrees. Clearly $C\cup D$ identifies a partition of the plane; The "perimeter" $P$ of $C\cup D$, the unbounded subset $X$ consisting of points "outside of" $P$, and the bounded subset $Y$ (possibly empty; see above) consisting of points "inside of" $P$. If the curve $C'$ is a subset of $X$, then it will not be "going between" $A$ and $B$ as it is supposed to, therefore it has a point shared with $P$, and we are done.
Case 2 $\quad$ The curve $C$ passes "above" $B'$ or "below" $A'$. We consider only the former, and the latter is similar. Again, we let $D$ be the rotation of $C$ by $180$ degrees, and we have $P$, $X$, and $Y$ as in case 1. In this case we know that $A'$ and $B'$ are in $Y$. If the curve $C'$ along with its $180$ degree rotation $D'$ want to identify, using their perimeter, a bounded subset of the plane that contains $A$ and $B$ (as they are supposed to), then they must "step outside" of $X$. And when they do so, they intersect $P$, and we are done again.
Below are five figures that help illustrate the idea:
The above can be formalized using concepts of elementary topology, and the not-so-elementary theorem known as Jordan Curve Theorem. I will try to post a formal answer once I know what to do with the cases I am missing.
For now, I will give a few formulations that can motivate a formalization.
By Jordan Curve Theorem, every closed curve ($C \cup D$ in our application) divides the plane into at least two path-connected components, one of which is the unbounded exterior which I call $X$. The union of the other bounded components is what I call $Y$, and the perimeter $P$ is the closure of $Y$ without $Y$'s interior.
As to what I mean by "below" or "above": Given a curve $C$, we walk on it from $A$ to $B$, and we take note how we travel by writing a sting $\rho$ of Greek letters. If we pass through the region of the $Y$-axis below $A'$, we write $\alpha$ for going left-to-right and $\alpha^{-1}$ for going right-to-left. Similarly, we use $\beta$ and $\gamma$ for the region between $A'$ and $B'$ and the region above $B'$, respectively. We use $\delta$ (or $\epsilon$, respectively) when we make a clockwise rotation around $A$ (or $B$), and $\delta^{-1}$ (or $\epsilon^{-1}$) for a counterclockwise turn. Then we verify that Case 1 deals with those curves whose reduced $\rho = \sigma \beta \tau$, where $\sigma$ (or $\tau$, respectively) only refers to rotations about $A$ (or $B$). While Case 2 deals with those where $\rho = \sigma \alpha \tau$ or $\sigma \delta \tau$, where $\sigma$ and $\tau$ are as before. Furthermore, the situations which I miss are those whose $\rho$ does not have such a form.