Suppose we have $p,q\in\mathbb{Z}':=\mathbb{Z}\cup\left\{-\infty,+\infty\right\}$ and consider the inverse stereographic projection on the open semi-circle as shown in the picture below, that is $p$ is mapped to $p'$ and $q$ is mapped to $q'$.
We define the metric $d$ on $\mathbb{Z}'$ to be the arc length between $p'$ and $q'$, i.e.
$$ d(p,q):=\arccos\left(\frac{pq+1}{\sqrt{(p^2+1)(q^2+1)}}\right),~~~p,q\in\mathbb{Z}'. $$
(1) Since we are only considering $\mathbb{Z}'$ (and not $\mathbb{R}\cup\left\{-\infty,+\infty\right\}$), the inverse stereographic projection only gives "discrete points" on the semi-circle. Am I right that for $p\in\mathbb{Z}$ we always have the following implication: $$ d(p,q)<\min\left\{d(p,p-1),d(p,p+1)\right\}\implies p=q? $$
(That is if the distance between $p$ and $q$ is smaller than the distance between $p$ and its neighbour points $p-1$ and $p+1$, we need to have $p=q$.)
(2) Is there something similar for $p\in\left\{\pm\infty\right\}$? Does there exist some $q\in\mathbb{Z}$ such that $$ d(p,+\infty)<d(q,+\infty)\implies p=+\infty? $$ I guess not since topologically each neighborhood of $+\infty$ contains some $$ \mathbb{Z}\cap (a,+\infty]:=\left\{x\in\mathbb{Z}: x>a\right\}\cup\left\{+\infty\right\} $$ and hence some $x\in\mathbb{Z}$ such that $d(x,+\infty)<d(q,+\infty)$ and $x<\infty$.
For your first point, I believe you are correct. One way to see this is to observe that for each fixed $p\in \mathbb Z$, the quotient $Q_p:\mathbb R\to \mathbb R$ given by \begin{equation*} Q_p(q) = \frac{pq+1}{\sqrt{(p^2 + 1)(q^2 + 1)}} \end{equation*} is strictly increasing for $q< p$ and strictly decreasing for $q>p$. In particular, for all integers $q$ satisfying $q\neq p$ we have \begin{equation*} Q_p(q)\leq \max (Q_p(p -1), Q_p(p + 1)). \end{equation*} Now run this through arccosine.
For your second point I believe you are also correct (no such $q\in \mathbb Z$ exists) and I agree with your reasoning. Another way to see that no such $q$ exists is to suppose $q\in \mathbb Z$ exists and apply the implication with $p= q +1$ as (the neighbor of $q$ whose distance to $+\infty$ is less than $d(q, +\infty)$). This gives $q + 1 = +\infty$, a contradiction.