Intro question:
For any $n$, Player 2 wins using a mirroring strategy of responding to a move $x$ with the antipodal move $-x$ (except in the corner case where $r>\sqrt 2$ where Player 1's first move removes everything and wins). Because each of Player's 2 move maintains the set of remaining points as symmetrical around the origin, as long as Player 1 has a valid move, so does Player 2. Since Player 2 can never lose and the game eventually ends, they win the game.
The strategy for the intro question suggests a generalized version that can apply to topological nim on different spaces.
For a metric space $X$ and radius $r$, say that a map $m$ from $X \to X$ is a mirror map if:
- $m$ is an isometry: it is preserves distances on $X$
- $m$ is self-inverse: $m^{-1}=m$
- $m$ moves each point at least distance $r$ away: $d(x, m(x)) \geq r$ for all $x\in X$
For the intro question, the mirror map is $m(x)=-x$.
We find that if a mirror map $m$ exists on $X$ with radius $r$, then Player 2 wins topological nim. They do this by responding to any move $x$ of of Player 1 with $m(x)$, maintaining the $m$-symmetry of the remaining space.
Because $d(x, m(x)) \geq r$, the point $m(x)$ is not removed following $x$ and so remains a legal move. Moreover $m(x)$ cannot have been removed by any earlier move $y$ or $m(y)$ because $m$ is distance-preserving and self-inverse: if $d(m(x),y)<r$ then $d(x,m(y))<r$, so $x$ would already have been an illegal move. Similarly if $d(m(x),m(y))<r$, then $d(x,y)<r$.
We can apply this general argument to the main problem to first show that:
Player 2 wins when $n$ is odd, except in the trivial case $r > \sqrt{2}$ when Player 1's first move removes everything and wins.
For $\mathbb{RP}^n$ with $n+1=2k$ even, the mirror map $m$ that rotates each adjacent pair of coordinates 90 degrees as $(x,y) \to (y,-x)$, which is self-inverse because a 180-degree rotation is the identity. That is, $m$ sends $(x_1,y_1,x_2,y_2,\dots,x_k,y_k)$ to $(y_1,-x_1,y_2,-x_2,\dots,y_k,-x_k)$. It's easy to see that $m$ is an isometry.
The Euclidian distance-squared from $(x,y)$ to $(y,-x)$ is always $2$, and likewise to its antipode $(-y,x)$. So the total distance-squared $d(x,m(x))^2$ is twice the sum of the coordinates-squared, which is $2$ because the points lie on a unit sphere. So, this satisfies the distance property of a mirror map as long as $r \leq \sqrt 2$. If $r>\sqrt 2$, then Player 1's first move trivially wins by removing everything.
For the other case, we can draw inspiration from the strategy for a modified "intro question", where the players play on a disk of radius $1$ in $\mathbb{R}^2$. For this game:
Player 1 wins by placing their first move in the center point $p=0$, then continuing to a mirror strategy of $m(x)=-x$. After their first move, Player 1 takes on the role of the new Player 2 to win with the mirror strategy on the remaining space. Note that while $m(x)=-x$ is not a mirror map on the original space due to points near the origin remaining near the origin, the first move removes such points, and ensures that no play interferes with its mirror.
This same type of argument lets us handle the other case (edit: this doesn't work, see Jaap Scherphuis's comment):
Player 1 wins when $n$ is even.
Player 1 starts by moving to $p=\pm (1,0,0 \dots)$, or really to anywhere and rotating their coordinates appropriately. They play on the remaining space as the new second player using mirror map $m$ like before of applying a 90 degree rotation $(x,y) \to (y,-x)$ on each pair of adjacent coordinates, except the unpaired first coordinate remains unchanged. That is, $m$ sends $(a, x_1,y_1,x_2,y_2,\dots,x_k,y_k)$ to $(a, y_1,-x_1,y_2,-x_2,\dots,y_k,-x_k)$. (Recall that the total number of coordinates $2k+1=n+1$ is odd.) This map is still distance-preserving and self-inverse as before.
The intuition here is that after removing the "ice caps" near the poles $p= \pm (1,0,0 \dots)$, the remaining space is all sufficiently far-moved by a rotation $m$ whose fixed points are the poles. We will show that the distance moved by the 90-degree rotation is at least the distance to the pole $p$: $d(x,m(x)) \geq d(p,x)$. This implies that as long as $x$ was a legal move, so is $m(x)$.
We can express $d(x,m(x))^2$ by looking at the sum of the coordinate differences squared, which is $2(x_1^2+y_1^2+\cdots+x_k^2+y_k^2)=2(1-a^2)$, so we have $d(x,m(x))^2= 2(1-a^2)$. (Flipping $a$ to $-a$ only gives a bigger distance). Similarly looking coordinate-wise, and assuming WLOG that $a \geq 0$, we have the distance squared to the initial move is $d(x,p)^2 =(1-a)^2 + 1-a^2$, which is smaller or equal on the interval $a \in [0,1]$ as desired.
