There is a Theorem in the book that says:
The space $\mathbb{C}P^n$ is CW complex of dimension $2n$.
I wonder some questions:
Is there any Theorem or result that if a space has CW complex structure, then it is a manifold?
Furthermore, to prove that it is of dimension $2n$, can one use a homology group of it? Is there any relation between dimension and homology group?
PS. Please do not answer using tools in Differential Geometry or Differential Topology.
No, having a CW complex structure is absolutely not enough to be a manifold. Being a CW complex is very easy, being a manifold is hard. Take the wedge sum of two circles $S^1 \vee S^1$ for example: it's a CW complex, but not a manifold.
If a space $M$ is a compact manifold without boundary, then you can read its dimension using singular homology: it is the greatest integer $n$ such that $H_n(M;\mathbb{Z}/2\mathbb{Z})$ is nonzero (and then it will be equal to $\mathbb{Z}/2\mathbb{Z}$). But this requires 1/ to know that $M$ is a manifold (without boundary) and 2/ to know that it is compact. Indeed, look at the homology of $\mathbb{R}^n$ and try to read its dimension there.
In the previous paragraph, the word "dimension" referred to the dimension of a manifold: the dimension of $M$ is the number $n$ such that $M$ is locally homeomorphic to $\mathbb{R}^n$. The dimension of a CW complex is something else: it is the maximum dimension of the cells of the CW complex. If $X$ is a CW complex and you know it's a manifold, then the two notions agree, but in general you can talk about the dimension of a CW complex without knowing that its a manifold and vice-versa.
For your particular problem, the most straight-forward way is simply to give an explicit CW decomposition of $\mathbb{CP}^n$. This will prove that it's a CW complex, and since you will only have used cell of dimension $\le 2n$ it will prove it has dimension $2n$. To prove that it is a manifold is a different question altogether.