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
Looks good so far!
Another approach is to observe the space can be expressed as $S = X \cup Y$ where the subspaces $X$ and $Y$ are copies of the real line hence second countable. There should be a theorem somewhere that says the union of two second countable subspaces is itself second countable.