I'm trying to work on proving the following statement:
Let $\Bbb P^n$ be the $n$-dimensional (real) projective space. Then $\Bbb P^n$ has a CW decomposition with one cell in each dimension $0,...,n$, s.t. the $k$-skeleton is $\Bbb P^k$ for $0<k<n$.
Proceeding by induction on $n$ seemed to be the most natural. For $k=1$, the space $\Bbb P^1\subseteq\Bbb R^2$ has a CW decomposition consisting of the $1$-cell $\Bbb R$. Let's assume we have constructed the spaces $\Bbb P^j$ for all $1\le j\le n-1$. We can realize $\Bbb P^k$ as subspace of $\Bbb P^n$ through the usual inclusion $\Bbb R^{k+1}\hookrightarrow\Bbb R^{n+1}$ whenever $k<n$. Now, the map $F:\overline{\Bbb B^n}\to\Bbb R^{n+1}\setminus\{0\}$ defined by $\mathrm{x}\mapsto\left(\mathrm x, \sqrt{1-|\mathrm x|^2}\right)$ is continuous. Composing this map with the quotient map $q:\Bbb R^{n+1}\setminus\{0\}\to\Bbb P^n$ defining $\Bbb P^n$, we obtain the continuous map $q\circ F:\overline{\Bbb B^n}\to\Bbb P^n$, which serves as the desired characteristic map for an $n$-cell.
I have some skepticisms about the above argument. Firstly, we never used a $0$-cell as required. In fact, I was wondering how a $0$-cell could fit into the picture here since $\Bbb P^n$ is the set of all $1$-dim. linear subspaces of $\Bbb R^{n+1}$. Secondly, is the inductive step correct? I think I'm missing something to finish the argument.
First as you say $\mathbb{R} \mathbb{P}^{n-1} \hookrightarrow \mathbb{R} \mathbb{P}^n$ is a closed subspace, where the inclusion is given by $[x_0 : ... : x_{n-1}] \rightarrow [x_0 : ... : x_{n-1} : 0]$.
This corresponds to the equatorial part of $\mathbb{S}^n \subset \mathbb{R}^{n+1}$ (with all antipodes identified).
When $x_n \neq 0$, take the upper hemisphere of $\mathbb{S}^n$ as a coordinate chart, which is homeomorphic to an $n$-cell, $\mathbb{D}^n$ (its the graph of the map you gave on $\mathbb{D}^n$), the boundary of the upper hemisphere is the equatorial part, which after identifying antipodes corresponds to $\mathbb{R} \mathbb{P}^{n-1}$, so this tells you exactly how to glue this $n$-cell to $\mathbb{R} \mathbb{P}^{n-1}$, that is, the attaching map is just the quotient map $\partial \mathbb{D}^{n} = \mathbb{S}^{n-1} \rightarrow \mathbb{R} \mathbb{P}^{n-1}$ which identifies antipodal points.
Start with $\mathbb{R} \mathbb{P}^{0}$ given by $\partial [-1,1] / \sim$, where $\sim$ identifies $-1$ and $1$, in other words, $\mathbb{R} \mathbb{P}^{0}$ is just a point or a zero cell, and then use the inductive construction I outlined already to build $\mathbb{R} \mathbb{P}^1$ and so on, yielding one cell in every dimension.