$\mathbb{S}^1$-action on $\mathbb S^{2n+1}$ I am trying to show $\mathbb{C}P^n$ is Hausdorff. I came across a proof that I would like to understand.

Let $[z],[w] \in \mathbb{C}P^{n}$ such that $[z]\ne [w]$ and $z,w \in \mathbb{S}^{2n+1}\subset \mathbb{C}^{n+1}$. Note that $\mathbb{S}^1 \subset \mathbb{C}$ is compact so we define $R = \min \{|sz - tw|: s,t \in \mathbb{S}^1\subset \mathbb{C}\} > 0$. Let $U = \{x \in \mathbb{S}^{2n+1}:|x-z| < \frac{R}{2}\}$ and $V = \{x \in \mathbb{S}^{2n+1}:|x-w| < \frac{R}{2}\}$. Note that $\lambda U = U$ and $\lambda V = V$ for all $\lambda \in \mathbb{S}^1$. So it follows that $\pi^{-1}(\pi(U)) = U$ and $\pi^{-1}(\pi(V))=V$. Thus, $\pi(U)$ and $\pi(V)$ are open and are disjoint, respectively containing $[z]$ and $[w]$. It follows that $\mathbb{C}P^n$ is Hausdorff.

Note that $\pi$ refers to the quotient map. I do not understand why both $U$ and $V$ are invariant under that action of $\mathbb{S}^1$, and to be honest, I do not understand the the aim behind defining $R$ as they have.

This is false. The scalar $-1$ is in $\mathbb{S}^1$, but of every $z\in\mathbb{S}^{2n+1}$, $ |z - (-1)z| = 2$. Thus, if $\varepsilon = 1$ for example, $U$ is not stable under the action of $\mathbb{S}^1$.
