https://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox#Step_3
In the part of the proof that deals with the fixed points on $S^2$, the page says "Let $\lambda$ be some line through the origin that does not intersect any point in $D$. This is possible since $D$ is countable." where $D$ is the set of the fixed points. Could you formally explain why we can find such line that does not intersect any point in $D$? My understanding is that since $S^2$ is a uncountable set, there are infinitely many points on $S^2$ that are not in $D$. But I don't know how to explain such line through origin can intersect with $S^2$ on "two" points that are not in %D%.
The next line of the proof is "Let $J$ be the set of angles, $\alpha$, such that for some natural number $n$, and some $P$ in $D$, $r(n\alpha)P$ is also in $D$, where $r(n\alpha)$ is a rotation about $\lambda$ of $n\alpha$. Then $J$ is countable." Exactly why is $J$ countable?
Thank you in advance!
For the first part, I will repeat my answer from comments above - the point is that there are also uncountably many possible lines going through (at least) one point not in D, since (informally) "removing countably many points from an uncountable set leaves uncountably many points remaining". So if every such line also went through a point in D, this would imply that D is uncountable.
For the second part, we first fix a point $P$ of $D$ (if we can show the set of rotations about $P$ taking $P$ to another point of $D$ is countable, since $D$ is countable we will be done). Denote by $J_P$ the subset of $J$ of the angles $\alpha$ corresponding to rotations from $P$. Well, we know there are only countably many points in $D$, so in particular there are only countably many angles $\beta$ which we can rotate $P$ by to obtain another point of $D$. Now, each $\beta$ is in $J_P$, but we also have $$\beta = 2 \cdot (\beta/2) = 3 \cdot (\beta/3) = \ ... \ ,$$ so in fact each $\alpha_{\beta,n} := \beta/n$ is in $J_P$ too. But for each $\beta$, the set of $\alpha_{\beta,n}$ is countable, and I claim that $J_P = \{ \alpha_{\beta,n} \}$; indeed, if $\alpha$ is a rotation in $J_P$, then $n \alpha = \beta$ for some $n$ and angle $\beta$ as above. Thus $J_P$ is in particular countable (we can index this set by $(\beta,n) \in \{\text{these countably many angles} \ \beta \} \times \mathbb{Z}$, a countable set).
Since $J_P$ is countable for each $P$, and $D$ is countable, we deduce that $J$ is countable. More formally, we can index $D$ as $D = \{P_i \}_{i} = \{P_1, P_2, \ ... \}$. Then $J = J_{P_1} \cup J_{P_2} \cup \ ...$ is countable, since (assuming the axiom of choice) the countable union of countable sets is countable (see: https://proofwiki.org/wiki/Countable_Union_of_Countable_Sets_is_Countable).
Hope this makes sense! Let me know if you have any questions.