TITLE: Is there any example of space not having the homotopy type of a CW-complex?
QUESTION [16 upvotes]: What is an example of space not having the homotopy type of a CW-complex? Is there any general method that can prove that the given space does not have the homotopy type of a CW-complex?
(added) It would be more interesting if one can suggest a connected example.

REPLY [4 votes]: Here are two more examples similar to Mariano's answer (together with Moishe's comment).
Example 1. Let $A \subset \mathbb{R}$ be
$$A = \{ \frac{1}{n} \mid n \in \mathbb{N} \} \cup \{0\}.$$
Let $f \colon A \to Y$ be a map to a CW complex $Y$. Then the point $f(0) \in Y$ has a path-connected neighborhood $V \subseteq Y$, and there is a neighborhood $U$ of $0 \in A$ satisfying $f(U) \subseteq V$. Hence, $f$ induces a map on path components $f_* \colon \pi_0(A) \to \pi_0(Y)$ which is not injective. In particular, $f$ is not a homotopy equivalence.
Example 2. Let $T \subset \mathbb{R}^2$ be the topologist's sine curve
$$T = \{ (x, \sin \frac{1}{x}) \mid x \in (0,1] \} \cup \{(0,0)\}.$$
Let $f \colon T \to Y$ be a map to a CW complex $Y$. Since $T$ is connected, $f(T)$ is connected. Since the path components of $Y$ agree with its connected components, $f(T)$ lies within one path component of $Y$. Hence, $f$ induces a map on path components $f_* \colon \pi_0(T) \to \pi_0(Y)$ which is not injective.
Remark. Example 2 holds more generally for any space $X$ with some connected component that contains more than one path component. In fact, the property that connected components agree with path components is invariant under homotopy equivalence.