Is it true that a retract of a CW-complex is a CW-complex ?
I failed to find a counterexample! I was wondering if such spaces have any topological characterization ?
My second question is the following:
Suppose that $X$ is a CW complex and $Y$ is a space such that $X\times Y$ is a CW-complex. Is it true that $Y$ is also a CW-complex ?
The answer to your first question is "no". Here is a counterexample.
Let $C \subset [0,1]$ be the standard middle-thirds Cantor set.
Let $\{I_n\}_{n \in \mathbb{Z}}$ be an enumeration of the components of $[0,1] - C$, a pairwise disjoint set of open intervals, and let $I_n = (a_n - \epsilon_n,a_n + \epsilon_n)$.
In $\mathbb{R}^2$, let $D_n$ be the closed 2-dimensional disc with center $(a_n,0)$ and radius $\epsilon_n$.
Let $X = \bigl([0,1] \times \{0\}\bigr) \cup \bigl(\bigcup_{n \in \mathbb{Z}} D_n\bigr)$.
This space $X$ is a deformation retract of $\mathbb{R}^2$, but it is not a CW complex, because the closure of the set of points of $X$ that separates $X$ is $C \times \{0\}$, but the closure of the set of separating points of a CW complex is a subcomplex of the $1$-skeleton, and $C$ is not a 1-dimensional CW complex.