By definition,
$$
\mathbb{RP}^n = \left(\mathbb{R}^{n+1} \setminus \{0\}\right) \big/ \sim\, ,
$$
where
$$
x \sim y \iff \exists\ \lambda \in \mathbb{R} \setminus \{ 0 \} \text{ such that } x = \lambda y.
$$
The topology on $\mathbb{RP}^n$ is, by definition, the quotient topology induced by the canonical projection
\begin{align}
\pi &: \mathbb{R}^{n+1} \setminus \{ 0 \} \to \mathbb{RP}^n\\
&: (x_0,\dots,x_n) \mapsto [x_0,\dots,x_n]
\end{align}
where $[x_0,\dots,x_n] \in\mathbb{RP}^n$ denotes the equivalence class of $(x_0,\dots,x_n) \in \mathbb{R}^{n+1} \setminus \{0 \}$. This makes $\pi$ a quotient map.
To show that $\mathbb{RP}^n$ is locally Euclidean, we need to exhibit a cover for $\mathbb{RP}^n$ by coordinate charts. For each $0 \leq i \leq n$, define $U_i \subset \mathbb{R}^{n+1} \setminus \{ 0 \}$ by
$$
U_i = \left\{ (x_0,\dots,x_n) \in \mathbb{R}^{n+1} \setminus \{ 0 \} : x_i \neq 0 \right\}.
$$
One can check that $U_i$ is an open subset of $\mathbb{R}^{n+1} \setminus \{ 0 \}$. Define $V_i \subset \mathbb{RP}^n$ to be $\pi(U_i)$. Then, one can check that $V_i$ is an open subset of $\mathbb{RP}^n$ and $\pi_i = \pi |_{U_i} : U_i \to V_i$ is also a quotient map. The sets $V_i$, $0 \leq i \leq n$, form an open cover of $\mathbb{RP}^n$.
We show that each $V_i$ is homeomorphic to $\mathbb{R}^n$ as follows. For each $0 \leq i \leq n$, define the map $\psi_i : V_i \to \mathbb{R}^n$ by
$$
\psi_i[x_0,\dots,x_n] = \left( \frac{x_0}{x_i},\dots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\dots,\frac{x_n}{x_i} \right).
$$
Continuity of $\psi_i$:
The map $\varphi_i = \psi_i \circ \pi_i : U_i \to \mathbb{R}^n$ is given by
$$
\varphi_i(x_0,\dots,x_n) = \left( \frac{x_0}{x_i},\dots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\dots,\frac{x_n}{x_i} \right).
$$
Since $\varphi_i$ is continuous, by the characteristic property of quotient maps $\psi_i$ is also continuous.
Bijectivity of $\psi_i$:
Note that $\psi_i$ is surjective because for every $(u_1,\dots,u_n) \in \mathbb{R}^n$, $[u_1,\dots,u_i,1,u_{i+1},\dots,u_n] \in V_i$ and $\psi_i([u_1,\dots,u_i,1,u_{i+1},\dots,u_n]) = (u_1,\dots,u_n)$. Note that every element in $V_i$ has a unique representative whose $i$th coordinate equals $1$. This fact easily implies that $\psi_i$ is injective.
Continuity of $\psi_i^{-1}$:
For each $0 \leq i \leq n$, consider the map $\theta_i : \mathbb{R}^n \to \mathbb{R}^{n+1} \setminus \{ 0 \}$ given by
$$
\theta_i(u_1,\dots,u_n) = (u_1,\dots,u_i,1,u_{i+1},\dots,u_n).
$$
Then, $\theta_i$ is continuous and its image is contained in $V_i$. One now checks that $\pi_i \circ \theta_i = \psi_i^{-1}$. So, $\psi_i^{-1}$ is continuous.
Hence, $\psi_i$ is a homeomorphism for each $0 \leq i \leq n$.
To show that $\mathbb{RP}^n$ is Hausdorff, choose $\tilde{x}$ and $\tilde{y}$, two distinct points in $\mathbb{RP}^n$.
If there exists $0 \leq i \leq n$ such that both points lie in $V_i$, then $\psi_i(\tilde{x})$ and $\psi_i(\tilde{y})$ are two distinct points in $\mathbb{R}^n$. Since $\mathbb{R}^n$ is Hausdorff, there exists a pair of disjoint open sets $A$ and $B$ with $\psi_i(\tilde{x}) \in A$ and $\psi_i(\tilde{y}) \in B$. Hence, $\psi_i^{-1}(A)$ and $\psi_i^{-1}(B)$ are disjoint open subsets of $V_i$ (and hence of $\mathbb{RP}^n$) such that $\tilde{x} \in \psi_i^{-1}(A)$ and $\tilde{y} \in \psi_i^{-1}(B)$.
On the other hand, suppose there is no $i$, $0 \leq i \leq n$, such that $\tilde{x}$ and $\tilde{y}$ both lie in $V_i$. Let $(x_0,\dots,x_n)$ and $(y_0,\dots,y_n)$ be representatives of $\tilde{x}$ and $\tilde{y}$, respectively. There exists $i \neq j$, $0 \leq i,j \leq n$, such that
\begin{align}
&x_i \neq 0, y_i = 0, \quad \text{ and}\\
&x_j = 0, y_j \neq 0.
\end{align}
Fix the representatives so that $x_i = 1 = y_j$. WLOG, let $i < j$. Choose $0 < \epsilon < 1$. The sets
\begin{align}
A &= \{ [a_0,\dots,a_{i-1},1,a_{i+1},\dots,a_n] : |a_k - x_k| < \epsilon\ \forall\ k \neq i \} \subset V_i, \quad \text{ and}\\
B &= \{ [b_0,\dots,b_{j-1},1,b_{j+1},\dots,b_n] : |b_k - y_k| < \epsilon\ \forall\ k \neq j \} \subset V_j
\end{align}
are open sets containing $\tilde{x}$ and $\tilde{y}$, respectively. This is because $\psi_i(A)$ is an open rectangle in $\mathbb{R}^n$ centered on $\psi_i(\tilde{x})$ having side length $2 \epsilon$, and similarly $\psi_j(B)$ is an open rectangle in $\mathbb{R}^n$ centered on $\psi_j(\tilde{y})$ having side length $2 \epsilon$. They are disjoint because if $[a_0,\dots,a_{i-1},1,a_{i+1},\dots,a_n] = [b_0,\dots,b_{j-1},1,b_{j+1},\dots,b_n]$, then we must have $a_j \neq 0$, $b_i \neq 0$, and $a_j b_i = 1$. But, $|a_j| < 1$ and $|b_i| < 1$, so this is not possible.
Hence, $\mathbb{RP}^n$ is Hausdorff, and so $\mathbb{RP}^n$ is an $n$-manifold.
Best Answer
$\mathbb{R}P^n$ is usually described as
i) lines through the origin in $\mathbb{R}^{n+1}$ (topology defined by the natural "closeness" of lines, say neighborhoods of a line are all lines within angle $\theta$ from the line
ii) quotient of $S^n$ by the antipodal map, like i) with each line represented as two points which we identify
iii) equivalence classes of points in $\mathbb{R}^{n+1}\backslash \{0\}$, identify two points $x,y$ if $x=\lambda y, \lambda\in\mathbb{R}^{\times}$ which looks a lot like i) since the equivalence classes are lines through the origin (minus the origin)
the definition via the affine cover you give above is iii) broken into the pieces (using homogeneous coordinates) $U_i=\{[x_0:\cdots:x_{i-1}:1:x_{i+1}:\cdots:x_n]\ |\ x_0,...,\hat{x_i},...,x_n\in\mathbb{R}\}$ (just a note that $[x_0:\cdots:x_n]$ denotes the equivalence class of the line through $(x_0,...,x_n)$, called "homogeneous coordinates" for a point in projective space)