I think the comments answer the question, but to give you a reference:

Milnor, Stasheff: Characteristic Classes, Chapter 6

They prove that every Grasmann manifold $G_n(\mathbb{R}^m)$ is a CW-Complex. (The cells are constructed with Schubert symbols). The complex case works in the same fashion.
As a result you get that $\mathbb{CP}^n$ consists of $n+1$ cells: for every $0 \leq k \leq n$ you get one $2k$-cell. The $2k$-skeleton is a $\mathbb{CP}^k$

EDIT: Sorry for the sloppiness!
Not every CW-Complex is triangulable, but the following holds:
Every regular CW-Complex (and $\mathbb{CP}^n$ is a regular complex) $X$ is triangulable.
This is true, since the barycentric subdivision is a simplicial complex that is homeomorphic to $X$. For a full proof, see for example Cellular structures in topology (p.130) by Fritsch and Piccinini.

I think the comments answer the question, but to give you a reference:

Milnor, Stasheff: Characteristic Classes, Chapter 6

They prove that every Grasmann manifold $G_n(\mathbb{R}^m)$ is a CW-Complex. (The cells are constructed with Schubert symbols). The complex case works in the same fashion.
As a result you get that $\mathbb{CP}^n$ consists of $n+1$ cells: for every $0 \leq k \leq n$ you get one $2k$-cell. The $2k$-skeleton is a $\mathbb{CP}^k$

EDIT: Sorry for the sloppiness!
Not every CW-Complex is triangulable, but the following holds:
Every regular CW-Complex (and $\mathbb{CP}^n$ is a regular complex $\oplus$) $X$ is triangulable.
This is true, since the barycentric subdivision is a simplicial complex that is homeomorphic to $X$. For a full proof, see for example Cellular structures in topology (p.130) by Fritsch and Piccinini.

Edit 2: $\oplus$: Perhaps the next sloppiness: The CW-structure of $\mathbb{CP}^n$ obtained by Schubert cells isn't regular (the characteristic map is 2-to-1) but I think there exists a regular CW-structure. But this might be harder to prove than I thought?!

I think the comments answer the question, but to give you a reference:

Milnor, Stasheff: Characteristic Classes, Chapter 6

They prove that every Grasmann manifold $G_n(\mathbb{R}^m)$ is a CW-Complex. (The cells are constructed with Schubert symbols). The complex case works in the same fashion.
As a result you get that $\mathbb{CP}^n$ consists of $n+1$ cells: for every $0 \leq k \leq n$ you get one $2k$-cell. The $2k$-skeleton is a $\mathbb{CP}^k$

I think the comments answer the question, but to give you a reference:

Milnor, Stasheff: Characteristic Classes, Chapter 6

They prove that every Grasmann manifold $G_n(\mathbb{R}^m)$ is a CW-Complex. (The cells are constructed with Schubert symbols). The complex case works in the same fashion.
As a result you get that $\mathbb{CP}^n$ consists of $n+1$ cells: for every $0 \leq k \leq n$ you get one $2k$-cell. The $2k$-skeleton is a $\mathbb{CP}^k$

EDIT: Sorry for the sloppiness!
Not every CW-Complex is triangulable, but the following holds:
Every regular CW-Complex (and $\mathbb{CP}^n$ is a regular complex) $X$ is triangulable.
This is true, since the barycentric subdivision is a simplicial complex that is homeomorphic to $X$. For a full proof, see for example Cellular structures in topology (p.130) by Fritsch and Piccinini.