I am aware of a few analytical calculations showing that the stereographic projection sends circles on the sphere to circles on the equatorial plane. There are related questions here.
What about a geometrical proof? I have found one somewhere that goes as follows.
Let $O$ be the center of the sphere and $S$ the pole doing the projection. Choose a circle $C$ on the sphere lying on a plane $P$, and denote its center $A$. The rays joining $S$ to the circle $C$ cross the equatorial plane $P'$ at some points, which we want to prove form another circle. Let $M$ and $N$ be the two points on the circle $C$ such that they belong to the $SOA$ plane. Their stereographic projection on the equatorial plane are denoted by $M'$ and $N'$.
The angles noted in the attached figure can easily be proven equal, leading to the two angles $SMN$ and $SN'M'$ being equal. This means that the planes $P$ and $P'$ are symmetrical to each other by reflection across the axis $SA$ (shown as an arrow). Now, the proof follows saying that the "cone" from $S$ to $C$ has the same symmetry, so that its intersections with the two planes $P$ and $P'$ are "equivalent". As a consequence, the two intersections are circles, and this proves the theorem.
But wait. To me, this is not really a "cone". But fine, it does not need to be. However, it does not seem symmetrical across $SA$: taking the symmetric point of $N$ or $M$ will clearly not land on the same "cone". So that would mean that the two shapes made by intersecting with $P$ and $P'$ should not be equivalent; consequently, stereographic projection would not send circles to circles.
Where is my error?
The cone is perhaps an oblique one, but it is a cone we are looking at: based on the circle $C$ from plane $P$, with summit $S$.
(Note however, that being oblique only means, from other perspective, that it's a straight cone based on another conic section.)
Now the main point is that - as seen in current picture - the plane $P'$ meets this cone in the same angle as $P$ does. So the conic section of $P'$ will be similar to that of $P$, so will be a circle.
The wording is not too lucky: I guess the symmetry is about the cone itself.