How to show this property of $\mathbb{RP}^{n}$.

$\mathbb RP^n$ is known to be a Hausdorff space (it is even a manifold, but we do not need this fact here). The quotient map $p : \mathbb R^{n+1} \setminus \{0\} \to \mathbb RP^n$ restricts to a surjective map $q : S^n \to \mathbb RP^n$. For each $y \in \mathbb RP^n$, $q^{-1}(y)$ is a pair of antipodal points in $S^n$. Since $S^n$ is compact and $\mathbb RP^n$ is Hausdorff, $q$ is closed map and therefore a quotient map. This gives an alternative description of $\mathbb RP^n$: It is the quotient space obtained from $S^n$ by identifying antipodal points.

The map $q$ is also an open map: If $U \subset S^n$ is open, then $q^{-1}(q(U)) = U \cup (-U)$ is open, thus $q(U)$ is open in $\mathbb RP^n$. Note that $-U = a(U)$, where $a$ is the antipodal map on $S^n$ which is a homeomorphism.

Each of the sets $U_i = \{(x_0,\ldots,x_n) \in S^n \mid x_i > 0 \}$, $i=0,\ldots,n$, is open in $S^n$ and is mapped by $q$ bijectively onto $q(U_i)$. Since $q$ is an open map, $U_i$ is mapped by $q$ homeomorphically onto $q(U_i)$. Thus also each subset $E \subset U_i$ is mapped by $q$ homeomorphically onto $q(E)$.

Now let $E^i = \{(x_0,\ldots,x_n) \in S^n \mid x_i > 0, x_{i+1} = \ldots = x_n = 0 \} \subset U_i$. The $E^i$ are mapped by $q$ homeomorphically onto $A^i = q(E^i)$. Each $E^i$ can be identified with the upper open half-sphere of the $i$-sphere $S^i \subset \mathbb R^{i+1}$ and is therefore homeomorphic to the open unit ball in $\mathbb R^i$. Thus $A^i \approx E^i \approx \mathbb R^i$.

The $A^i$ are pairwise disjoint: If $x = (x_0,\ldots,x_n) \in E^i$ and $y = (y_0,\ldots,y_n) \in E^j$ for $i < j$, then $x_j = 0$ and $y_j > 0$, thus $x$ and $y$ are not antipodal and therefore $q(x) \ne q(y)$.

$\bigcup_{i=0}^n A^i = \mathbb RP^n$: Let $y = q(x)$ with $x =(x_0,\ldots,x_n)$. Let $i$ be the biggest index such that $x_i \ne 0$. We may assume that $x_i > 0$, otherwise we replace $x$ by $-x$ which also satisfies $q(-x) = y$. But then we see that $x \in E^i$, i.e. $y \in A^i$.