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The proofs that any two CW complexes that are K(G,n) are homotopy equivalent has no obvious extension to the general case. I assume it is known to be false in general, but I can't find an example.

Connor Malin
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The Hawaiian earring (which is a a compact path connected and locally path connected subset of the plane) is a $K(G,1)$. See https://en.wikipedia.org/wiki/Hawaiian_earring, https://mathoverflow.net/questions/163847/are-the-higher-homotopy-groups-of-the-hawaiian-earring-trivial. Its fundamental group $G$ is uncountable.

Another example is the Warsaw circle. It is $K(0,n)$ for all $n$. See How to show Warsaw circle is non-contractible? and Connor Malin's answer.

Paul Frost
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For a nice example consider the pseudocircle (https://en.wikipedia.org/wiki/Pseudocircle). This is a finite topological space which is weakly equivalent to the circle $S^1$. It follows that the psuedocircle is a $K(\mathbb{Z},1)$. However, any finite CW complex must be discrete, so it obviously cannot be CW.

Tyrone
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  • As far as I can tell this doesn't anything about the homotopy type, but it is an easy extension of what you say, as any map from the pseudocircle to a Hausdorff space is constant. –  Jan 13 '19 at 15:39
  • @MikeMiller, the Pseudocircle $X$ has the weak homotopy type of $S^1\simeq K(\mathbb{Z},1)$. For the reason you state it clearly cannot by homotopy equivalent to $S^1$, and cannot even admit a weak equivalence in the direction $X\rightarrow S^1$. – Tyrone Jan 13 '19 at 17:29
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    Yes, I understand :) I was commenting only because the OP asked for something not homotopy equivalent to a CW complex, and your last sentence is up to homeomorphism. It is a good answer. –  Jan 13 '19 at 17:32
  • @MikeMiller, thanks for pointing that out, I do agree that my statement is a bit ambiguous. I actually meant that if $Y$ is finite and discrete then necessarily $\pi_0Y=|Y|$ and $\pi_1Y=0$, so since $\pi_1X\cong\mathbb{Z}$ it cannot be discrete, and hence cannot be CW. I barely even remember what a homeomorphish is anymore ;). – Tyrone Jan 13 '19 at 17:42
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I think something like this might work: take a noncontractible space that has all homotopy groups zero (such a thing is necessarily not a CW complex, I think an example can be obtained from wedging two comb spaces) and wedge it with K(G,n). The resulting space should not be homotopy equivalent to a CW complex and (perhaps depending on the noncontractible space) be a K(G,n).

Connor Malin
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    Or just taking G = 0 and a non-contractible space which is weakly contractible gives a counterexample! –  Jan 12 '19 at 22:43