I'm trying to prove that reflecting a point about the x and y axes is equivalent to reflecting it about the origin. Is my proof valid? How could I improve it?
Proof:
Take a point $a$ in the first quadrant (without loss of generality).
Draw a line $l_1$from the origin to $a$ and let $\theta$ be the angle formed by $l_1$ and the x axis.
Reflect $a$ about the $x$ axis and call that point $b$.
Draw the line $l_2$ from the origin to $b$ and call $\beta$ the angle formed by $l_2$ and the $x$ axis, and call $\rho$ the angle formed by $l_2$ and the $y$ axis.
Then $\theta = \beta$ and $l_1 = l_2$.
Now reflect $b$ about the $y$ axis and call that point $c$.
Draw $l_3$ from the origin to $c$ and call $\phi$ the angle formed by $l_3$ and the $y$ axis.
Then $l_2 = l_3$ and therefore $l_3 = l_1$. Also, $\rho = \phi$.
Since the $x$ and $y$ axes are orthogonal, $\theta$ and $\beta$ are the complements of $\phi$ and $\rho$, therefore $\theta + \beta + \phi + \rho = 180$ degrees, and so $l_1 + l_3$ is the diameter of a circle with the origin as center.
Therefore $c$ is symmetric to $a$ with respect to the origin.
Sure. Without loss of generality, assume your original point is $z$ in the first quadrant. I will describe the steps on the left and the corresponding transformation they induce to the right
$$z\rightarrow \bar{z}\text{ (reflect through horizontal axis)}$$ $$\bar{z}\rightarrow \bar{z}i\text{ (rotate plane counteclockwise by 90 degrees)}$$ $$\bar{z}i\rightarrow \overline{\bar{z}i}\text{ (reflect again through horizontal)}$$ $$\overline{\bar{z}i}\rightarrow -i(\overline{\bar{z}i})\text{ (rotate back clockwise by 90 degrees)}$$
The composition of all transformations then is equivalent to what you get when simplifying the last line above, which is:
$$f(z)=-i(\overline{\bar{z}i})=-z$$
And this is precisely symmetry with respect to the origin.
Note also that the transformation is an involution, since:
$$f(f(z))=z$$