When showing that $\mathbb{R}P^n$ is a topological manifold, the atlas is given by charts $\varphi_i\colon U_i\to\mathbb{R}^n$, where the $U_i$ are the classes $[x^1,\dots,x^{n+1}]$ with $x_i\neq 0$, and the coordinate charts are defined by $$ \varphi_i[x^1,\dots,x^{n+1}]=\left(\frac{x^1}{x^i},\dots,\frac{x^{i-1}}{x^i},\frac{x^{i+1}}{x^i},\dots,\frac{x^{n+1}}{x^i}\right) $$ and $$ \varphi_i^{-1}(u^1,\dots,u^n)=[u^1,\dots,u^{i-1},1,u^i,\cdots,u^n]. $$
But why is $\varphi_i^{-1}$ is continuous?
I can't get a good grasp on what an open sets in $\Bbb RP^n$ look like. I look at them by taking their representatives in $\Bbb R^{n+1}\setminus 0$, and then I know the set is open if all the lines through those representatives, minus $0$, which looks like two cones missing their vertex is open in $\Bbb R^{n+1}\setminus 0$, but it's proving difficult to show the preimage of an open set is open under $\varphi_i^{-1}$.
$\Bbb{R}P^n$ is (usually) equipped with the quotient topology induced by
$$ \psi : \Bbb{R}^{n+1} \setminus \{0\} \to \Bbb{R}P^n, x \mapsto [x_1, \dots, x_{n+1}]. $$
In particular, $\psi$ is continuous. Why is that helpful?
Additional question: why is $U_i$ an open set?