I'll prove something stronger. $A$, $B$, and $C$ don't need to be pairwise disjoint; we only need $A\cap B\cap C$ to be $\emptyset$.
Consider the map:
This is defined, since the denominators are never zero; remember that $A\cap B\cap C=\emptyset$.
By Borsuk-Ulam, there is a point $x\in S^2$ such that $f(x)=f(-x)$.
If the first (or second) coordinate of $f(x)=f(-x)$ is zero, then $x$ and $-x$ are both in $A$ (or $B$). This is impossible, since $A$ and $B$ don't contain antipodal points.
If the coordinates of $f(x)$ add up to $1$, then $d(\pm x,C)=0$, and both $x$ and $-x$ are in $C$. This is impossible, since $C$ doesn't contain antipodal points.
The only possibility is that the coordinates are nonzero and don't add up to $1$. This means that $x$ and $-x$ are both not in any of the three sets. That would imply that $x$ is not in $A$, $B$, $C$, $-A$, $-B$, or $-C$, which means that they do not cover $S^2$. QED
Note that we need the condition $A\cap B\cap C=\emptyset$. If the triple intersection is allowed to be nonempty, the theorem is false. For example, consider dividing the northern hemisphere into three equal parts, letting three lines of longitude (the vertical ones) be the borders between the sets. Then the three sets intersect at the north pole, and $A$, $B$, $C$ and their opposites cover the sphere.