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.