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Here are two examples similar to Ben's answer, followed by an example with a path-connected space.

Let $[X,Y]$ denote the naive pointed homotopy classes of pointed maps. We're comparing the (say, unreduced) singular cohomology functor $H^n(X;G)$ to the naively represented functor $[X_+, K(G,n)]$. As Chris pointed out, singular cohomology is representable in the derived sense, i.e., agrees with its value on a CW approximation of $X$.

Example 1. Let $T \subset \mathbb{R}^2$ be the topologist's sine curve $$T = \{ (x, \sin \frac{1}{x}) \mid x \in (0,1] \} \cup \{(0,0)\}.$$ Its zeroth singular cohomology is the finite product $$H^0(T;\mathbb{Z}) = \mathrm{Hom}_{\mathbb{Z}}(H_0(T;\mathbb{Z}), \mathbb{Z}) = \mathrm{Set}(\pi_0(T),\mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}.$$ Its zeroth naive cohomology is $$[T_+, K(\mathbb{Z},0)] = \mathrm{Set}(\ast,\mathbb{Z}) = \mathbb{Z}.$$ A CW approximation $S^0 \stackrel{\sim}{\to} T$ induces on zeroth naive cohomology the diagonal inclusion $\mathbb{Z} \hookrightarrow \mathbb{Z} \times \mathbb{Z}$.

Example 2. Let $X \subset \mathbb{R}$ be $$X = \{ \frac{1}{n} \mid n \in \mathbb{N} \} \cup \{0\}.$$ Let me name the points $x_n = \frac{1}{n}$ and $x_{\infty} = 0$ for convenience. The zeroth singular cohomology is the countable product $$H^0(X;\mathbb{Z}) = \mathrm{Set}(\pi_0(X),\mathbb{Z}) = \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z}.$$ The zeroth naive cohomology is the subgroup $$[X_+, K(\mathbb{Z},0)] = \{ (a_n) \in \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z} \mid (a_n) \text{ is eventually constant}\},$$ which happens to be a free abelian group on countably many generators. A CW approximation from the discrete space $\coprod_{x \in X} \{x\} = \mathbb{N} \cup \{\infty\} \stackrel{\sim}{\to} X$ induces on zeroth naive cohomology the inclusion of said subgroup.

Example 3. Let $E = \bigcup_{n \in \mathbb{N}} C_n$ be the Hawaiian earring, where $C_n \subset \mathbb{R}^2$ denotes the circle of radius $\frac{1}{n}$ centered at $(\frac{1}{n},0)$. There is a surjective map $H_1(E; \mathbb{Z}) \twoheadrightarrow \prod_{n \in \mathbb{N}} \mathbb{Z}$; see for instance Hatcher Example 1.25 or [1]. Dualization into $\mathbb{Q}$ yields an injective map $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) \hookrightarrow \mathrm{Hom}_{\mathbb{Z}}(H_1(E; \mathbb{Z}), \mathbb{Q}) = H^1(E; \mathbb{Q}),$$ which shows that the first singular cohomology $H^1(E; \mathbb{Q})$ has uncountable dimension as a vector space over $\mathbb{Q}$. Indeed, the vector space $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) = \mathrm{Hom}_{\mathbb{Q}}(\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z}), \mathbb{Q})$$ has uncountable dimension since $\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z})$ does.

Now denote $W_n = \bigcup_{i=1}^n C_i \cong \bigvee_{i=1}^n S^1$ and consider the retraction $E \to W_n$ that collapses all the circles $C_i$ with $i>n$ to the basepoint. Take a CW complex $K(\mathbb{Q},1)$. Any (pointed) map $E \to K(\mathbb{Q},1)$ is (pointed) homotopic to a map which factors as $E \to W_n \to K(\mathbb{Q},1)$ for $n$ large enough. This shows that the map $$\bigoplus_{i \in \mathbb{N}} \mathbb{Q} = \operatorname{colim}_n [W_n, K(\mathbb{Q},1)] \twoheadrightarrow [E,K(\mathbb{Q},1)]$$ is surjective, and thus the naive cohomology group $[E,K(\mathbb{Q},1)]$ is countable. In particular, a CW approximation $X' \stackrel{\sim}{\to} X$ induces a non-surjective map on naive cohomology $[E,K(\mathbb{Q},1)] \to H^1(E;\mathbb{Q})$.

Edit. For the record, I previously thought that a similar argument would work with integral cohomology $H^1(E;\mathbb{Z})$, but I had forgotten Specker's theorem $\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Z}) \cong \bigoplus_{i \in \mathbb{N}} \mathbb{Z}$.

In fact, it turns out that said cohomology group is $H^1(E;\mathbb{Z}) \cong \bigoplus_{i \in \mathbb{N}} \mathbb{Z}$, as explained here.

[1] Katsuya Eda and Kazuhiro Kawamura, MR 1772189 The singular homology of the Hawaiian earring, J. London Math. Soc. (2) 62 (2000), no. 1, 305--310.

Here are two examples similar to Ben's answer, followed by an example with a path-connected space.

Let $[X,Y]$ denote the naive pointed homotopy classes of pointed maps. We're comparing the (say, unreduced) singular cohomology functor $H^n(X;G)$ to the naively represented functor $[X_+, K(G,n)]$. As Chris pointed out, singular cohomology is representable in the derived sense, i.e., agrees with its value on a CW approximation of $X$.

Example 1. Let $T \subset \mathbb{R}^2$ be the topologist's sine curve $$T = \{ (x, \sin \frac{1}{x}) \mid x \in (0,1] \} \cup \{(0,0)\}.$$ Its zeroth singular cohomology is the finite product $$H^0(T;\mathbb{Z}) = \mathrm{Hom}_{\mathbb{Z}}(H_0(T;\mathbb{Z}), \mathbb{Z}) = \mathrm{Set}(\pi_0(T),\mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}.$$ Its zeroth naive cohomology is $$[T_+, K(\mathbb{Z},0)] = \mathrm{Set}(\ast,\mathbb{Z}) = \mathbb{Z}.$$ A CW approximation $S^0 \stackrel{\sim}{\to} T$ induces on zeroth naive cohomology the diagonal inclusion $\mathbb{Z} \hookrightarrow \mathbb{Z} \times \mathbb{Z}$.

Example 2. Let $X \subset \mathbb{R}$ be $$X = \{ \frac{1}{n} \mid n \in \mathbb{N} \} \cup \{0\}.$$ Let me name the points $x_n = \frac{1}{n}$ and $x_{\infty} = 0$ for convenience. The zeroth singular cohomology is the countable product $$H^0(X;\mathbb{Z}) = \mathrm{Set}(\pi_0(X),\mathbb{Z}) = \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z}.$$ The zeroth naive cohomology is the subgroup $$[X_+, K(\mathbb{Z},0)] = \{ (a_n) \in \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z} \mid (a_n) \text{ is eventually constant}\},$$ which happens to be a free abelian group on countably many generators. A CW approximation from the discrete space $\coprod_{x \in X} \{x\} = \mathbb{N} \cup \{\infty\} \stackrel{\sim}{\to} X$ induces on zeroth naive cohomology the inclusion of said subgroup.

Example 3. Let $E = \bigcup_{n \in \mathbb{N}} C_n$ be the Hawaiian earring, where $C_n \subset \mathbb{R}^2$ denotes the circle of radius $\frac{1}{n}$ centered at $(\frac{1}{n},0)$. There is a surjective map $H_1(E; \mathbb{Z}) \twoheadrightarrow \prod_{n \in \mathbb{N}} \mathbb{Z}$; see for instance Hatcher Example 1.25 or [1]. Dualization into $\mathbb{Q}$ yields an injective map $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) \hookrightarrow \mathrm{Hom}_{\mathbb{Z}}(H_1(E; \mathbb{Z}), \mathbb{Q}) = H^1(E; \mathbb{Q}),$$ which shows that the first singular cohomology $H^1(E; \mathbb{Q})$ has uncountable dimension as a vector space over $\mathbb{Q}$. Indeed, the vector space $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) = \mathrm{Hom}_{\mathbb{Q}}(\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z}), \mathbb{Q})$$ has uncountable dimension since $\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z})$ does.

Now denote $W_n = \bigcup_{i=1}^n C_i \cong \bigvee_{i=1}^n S^1$ and consider the retraction $E \to W_n$ that collapses all the circles $C_i$ with $i>n$ to the basepoint. Take a CW complex $K(\mathbb{Q},1)$. Any (pointed) map $E \to K(\mathbb{Q},1)$ is (pointed) homotopic to a map which factors as $E \to W_n \to K(\mathbb{Q},1)$ for $n$ large enough. This shows that the map $$\bigoplus_{i \in \mathbb{N}} \mathbb{Q} = \operatorname{colim}_n [W_n, K(\mathbb{Q},1)] \twoheadrightarrow [E,K(\mathbb{Q},1)]$$ is surjective, and thus the naive cohomology group $[E,K(\mathbb{Q},1)]$ is countable. In particular, a CW approximation $E' \stackrel{\sim}{\to} E$ induces a non-surjective map on naive cohomology $[E,K(\mathbb{Q},1)] \to H^1(E;\mathbb{Q})$.

Edit. For the record, I previously thought that a similar argument would work with integral cohomology $H^1(E;\mathbb{Z})$, but I had forgotten Specker's theorem $\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Z}) \cong \bigoplus_{i \in \mathbb{N}} \mathbb{Z}$.

In fact, it turns out that said cohomology group is $H^1(E;\mathbb{Z}) \cong \bigoplus_{i \in \mathbb{N}} \mathbb{Z}$, as explained here.

[1] Katsuya Eda and Kazuhiro Kawamura, MR 1772189 The singular homology of the Hawaiian earring, J. London Math. Soc. (2) 62 (2000), no. 1, 305--310.

Here are two examples similar to Ben's answer, followed by an example with a path-connected space.

Let $[X,Y]$ denote the naive pointed homotopy classes of pointed maps. We're comparing the (say, unreduced) singular cohomology functor $H^n(X;G)$ to the naively represented functor $[X_+, K(G,n)]$. As Chris pointed out, singular cohomology is representable in the derived sense, i.e., agrees with its value on a CW approximation of $X$.

Example 1. Let $T \subset \mathbb{R}^2$ be the topologist's sine curve $$T = \{ (x, \sin \frac{1}{x}) \mid x \in (0,1] \} \cup \{(0,0)\}.$$ Its zeroth singular cohomology is the finite product $$H^0(T;\mathbb{Z}) = \mathrm{Hom}_{\mathbb{Z}}(H_0(T;\mathbb{Z}), \mathbb{Z}) = \mathrm{Set}(\pi_0(T),\mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}.$$ Its zeroth naive cohomology is $$[T_+, K(\mathbb{Z},0)] = \mathrm{Set}(\ast,\mathbb{Z}) = \mathbb{Z}.$$ A CW approximation $S^0 \stackrel{\sim}{\to} T$ induces on zeroth naive cohomology the diagonal inclusion $\mathbb{Z} \hookrightarrow \mathbb{Z} \times \mathbb{Z}$.

Example 2. Let $X \subset \mathbb{R}$ be $$X = \{ \frac{1}{n} \mid n \in \mathbb{N} \} \cup \{0\}.$$ Let me name the points $x_n = \frac{1}{n}$ and $x_{\infty} = 0$ for convenience. The zeroth singular cohomology is the countable product $$H^0(X;\mathbb{Z}) = \mathrm{Set}(\pi_0(X),\mathbb{Z}) = \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z}.$$ The zeroth naive cohomology is the subgroup $$[X_+, K(\mathbb{Z},0)] = \{ (a_n) \in \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z} \mid (a_n) \text{ is eventually constant}\},$$ which happens to be a free abelian group on countably many generators. A CW approximation from the discrete space $\coprod_{x \in X} \{x\} = \mathbb{N} \cup \{\infty\} \stackrel{\sim}{\to} X$ induces on zeroth naive cohomology the inclusion of said subgroup.

Example 3. Let $E = \bigcup_{n \in \mathbb{N}} C_n$ be the Hawaiian earring, where $C_n \subset \mathbb{R}^2$ denotes the circle of radius $\frac{1}{n}$ centered at $(\frac{1}{n},0)$. There is a surjective map $H_1(E; \mathbb{Z}) \twoheadrightarrow \prod_{n \in \mathbb{N}} \mathbb{Z}$; see for instance Hatcher Example 1.25 or [1]. Dualization into $\mathbb{Q}$ yields an injective map $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) \hookrightarrow \mathrm{Hom}_{\mathbb{Z}}(H_1(E; \mathbb{Z}), \mathbb{Q}) = H^1(E; \mathbb{Q}),$$ which shows that the first singular cohomology $H^1(E; \mathbb{Q})$ has uncountable dimension as a vector space over $\mathbb{Q}$. Indeed, the vector space $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) = \mathrm{Hom}_{\mathbb{Q}}(\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z}), \mathbb{Q})$$ has uncountable dimension since $\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z})$ does.

Now denote $W_n = \bigcup_{i=1}^n C_i \cong \bigvee_{i=1}^n S^1$ and consider the retraction $E \to W_n$ that collapses all the circles $C_i$ with $i>n$ to the basepoint. Take a CW complex $K(\mathbb{Q},1)$. Any (pointed) map $E \to K(\mathbb{Q},1)$ is (pointed) homotopic to a map which factors as $E \to W_n \to K(\mathbb{Q},1)$ for $n$ large enough. This shows that the map $$\bigoplus_{i \in \mathbb{N}} \mathbb{Q} = \operatorname{colim}_n [W_n, K(\mathbb{Q},1)] \twoheadrightarrow [E,K(\mathbb{Q},1)]$$ is surjective, and thus the naive cohomology group $[E,K(\mathbb{Q},1)]$ is countable. In particular, a CW approximation $X' \stackrel{\sim}{\to} X$ induces a non-surjective map on naive cohomology $[E,K(\mathbb{Q},1)] \to H^1(E;\mathbb{Q})$.

[1] Katsuya Eda and Kazuhiro Kawamura, MR 1772189 The singular homology of the Hawaiian earring, J. London Math. Soc. (2) 62 (2000), no. 1, 305--310.

Here are two examples similar to Ben's answer, followed by an example with a path-connected space.

Let $[X,Y]$ denote the naive pointed homotopy classes of pointed maps. We're comparing the (say, unreduced) singular cohomology functor $H^n(X;G)$ to the naively represented functor $[X_+, K(G,n)]$. As Chris pointed out, singular cohomology is representable in the derived sense, i.e., agrees with its value on a CW approximation of $X$.

Example 1. Let $T \subset \mathbb{R}^2$ be the topologist's sine curve $$T = \{ (x, \sin \frac{1}{x}) \mid x \in (0,1] \} \cup \{(0,0)\}.$$ Its zeroth singular cohomology is the finite product $$H^0(T;\mathbb{Z}) = \mathrm{Hom}_{\mathbb{Z}}(H_0(T;\mathbb{Z}), \mathbb{Z}) = \mathrm{Set}(\pi_0(T),\mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}.$$ Its zeroth naive cohomology is $$[T_+, K(\mathbb{Z},0)] = \mathrm{Set}(\ast,\mathbb{Z}) = \mathbb{Z}.$$ A CW approximation $S^0 \stackrel{\sim}{\to} T$ induces on zeroth naive cohomology the diagonal inclusion $\mathbb{Z} \hookrightarrow \mathbb{Z} \times \mathbb{Z}$.

Example 2. Let $X \subset \mathbb{R}$ be $$X = \{ \frac{1}{n} \mid n \in \mathbb{N} \} \cup \{0\}.$$ Let me name the points $x_n = \frac{1}{n}$ and $x_{\infty} = 0$ for convenience. The zeroth singular cohomology is the countable product $$H^0(X;\mathbb{Z}) = \mathrm{Set}(\pi_0(X),\mathbb{Z}) = \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z}.$$ The zeroth naive cohomology is the subgroup $$[X_+, K(\mathbb{Z},0)] = \{ (a_n) \in \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z} \mid (a_n) \text{ is eventually constant}\},$$ which happens to be a free abelian group on countably many generators. A CW approximation from the discrete space $\coprod_{x \in X} \{x\} = \mathbb{N} \cup \{\infty\} \stackrel{\sim}{\to} X$ induces on zeroth naive cohomology the inclusion of said subgroup.

Example 3. Let $E = \bigcup_{n \in \mathbb{N}} C_n$ be the Hawaiian earring, where $C_n \subset \mathbb{R}^2$ denotes the circle of radius $\frac{1}{n}$ centered at $(\frac{1}{n},0)$. There is a surjective map $H_1(E; \mathbb{Z}) \twoheadrightarrow \prod_{n \in \mathbb{N}} \mathbb{Z}$; see for instance Hatcher Example 1.25 or [1]. Dualization into $\mathbb{Q}$ yields an injective map $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) \hookrightarrow \mathrm{Hom}_{\mathbb{Z}}(H_1(E; \mathbb{Z}), \mathbb{Q}) = H^1(E; \mathbb{Q}),$$ which shows that the first singular cohomology $H^1(E; \mathbb{Q})$ has uncountable dimension as a vector space over $\mathbb{Q}$. Indeed, the vector space $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) = \mathrm{Hom}_{\mathbb{Q}}(\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z}), \mathbb{Q})$$ has uncountable dimension since $\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z})$ does.

Now denote $W_n = \bigcup_{i=1}^n C_i \cong \bigvee_{i=1}^n S^1$ and consider the retraction $E \to W_n$ that collapses all the circles $C_i$ with $i>n$ to the basepoint. Take a CW complex $K(\mathbb{Q},1)$. Any (pointed) map $E \to K(\mathbb{Q},1)$ is (pointed) homotopic to a map which factors as $E \to W_n \to K(\mathbb{Q},1)$ for $n$ large enough. This shows that the map $$\bigoplus_{i \in \mathbb{N}} \mathbb{Q} = \operatorname{colim}_n [W_n, K(\mathbb{Q},1)] \twoheadrightarrow [E,K(\mathbb{Q},1)]$$ is surjective, and thus the naive cohomology group $[E,K(\mathbb{Q},1)]$ is countable. In particular, a CW approximation $X' \stackrel{\sim}{\to} X$ induces a non-surjective map on naive cohomology $[E,K(\mathbb{Q},1)] \to H^1(E;\mathbb{Q})$.

Edit. For the record, I previously thought that a similar argument would work with integral cohomology $H^1(E;\mathbb{Z})$, but I had forgotten Specker's theorem $\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Z}) \cong \bigoplus_{i \in \mathbb{N}} \mathbb{Z}$.

In fact, it turns out that said cohomology group is $H^1(E;\mathbb{Z}) \cong \bigoplus_{i \in \mathbb{N}} \mathbb{Z}$, as explained here.

[1] Katsuya Eda and Kazuhiro Kawamura, MR 1772189 The singular homology of the Hawaiian earring, J. London Math. Soc. (2) 62 (2000), no. 1, 305--310.

Here are two examples similar to Ben's answer, followed by an example with a path-connected space.

Let $[X,Y]$ denote the naive pointed homotopy classes of pointed maps. We're comparing the (say, unreduced) singular cohomology functor $H^n(X;G)$ to the naively represented functor $[X_+, K(G,n)]$. As Chris pointed out, singular cohomology is representable in the derived sense, i.e., agrees with its value on a CW approximation of $X$.

Example 1. Let $T \subset \mathbb{R}^2$ be the topologist's sine curve $$T = \{ (x, \sin \frac{1}{x}) \mid x \in (0,1] \} \cup \{(0,0)\}.$$ Its zeroth singular cohomology is the finite product $$H^0(T;\mathbb{Z}) = \mathrm{Hom}_{\mathbb{Z}}(H_0(T;\mathbb{Z}), \mathbb{Z}) = \mathrm{Set}(\pi_0(T),\mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}.$$ Its zeroth naive cohomology is $$[T_+, K(\mathbb{Z},0)] = \mathrm{Set}(\ast,\mathbb{Z}) = \mathbb{Z}.$$ A CW approximation $S^0 \stackrel{\sim}{\to} T$ induces on zeroth naive cohomology the diagonal inclusion $\mathbb{Z} \hookrightarrow \mathbb{Z} \times \mathbb{Z}$.

Example 2. Let $X \subset \mathbb{R}$ be $$X = \{ \frac{1}{n} \mid n \in \mathbb{N} \} \cup \{0\}.$$ Let me name the points $x_n = \frac{1}{n}$ and $x_{\infty} = 0$ for convenience. The zeroth singular cohomology is the countable product $$H^0(X;\mathbb{Z}) = \mathrm{Set}(\pi_0(X),\mathbb{Z}) = \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z}.$$ The zeroth naive cohomology is the subgroup $$[X_+, K(\mathbb{Z},0)] = \{ (a_n) \in \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z} \mid (a_n) \text{ is eventually constant}\},$$ which happens to be a free abelian group on countably many generators. A CW approximation from the discrete space $\coprod_{x \in X} \{x\} = \mathbb{N} \cup \{\infty\} \stackrel{\sim}{\to} X$ induces on zeroth naive cohomology the inclusion of said subgroup.

Example 3. Let $E = \bigcup_{n \in \mathbb{N}} C_n$ be the Hawaiian earring, where $C_n \subset \mathbb{R}^2$ denotes the circle of radius $\frac{1}{n}$ centered at $(\frac{1}{n},0)$. There is a surjective map $H_1(E; \mathbb{Z}) \twoheadrightarrow \prod_{n \in \mathbb{N}} \mathbb{Z}$; see for instance Hatcher Example 1.25 or [1]. Dualization yields an injective map $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Z}) \hookrightarrow \mathrm{Hom}_{\mathbb{Z}}(H_1(E; \mathbb{Z}), \mathbb{Z}) = H^1(E; \mathbb{Z}),$$ which shows that the first singular cohomology $H^1(E; \mathbb{Z})$ is not countably generated. Now denote $W_n = \bigcup_{i=1}^n C_i \cong \bigvee_{i=1}^n S^1$ and consider the retraction $E \to W_n$ that collapses all the circles $C_i$ with $i>n$ to the basepoint. Any (pointed) map $E \to K(\mathbb{Z},1) = S^1$ is (pointed) homotopic to a map which factors as $E \to W_n \to S^1$ for $n$ large enough. This shows that the map $$\bigoplus_{i \in \mathbb{N}} \mathbb{Z} = \operatorname{colim}_n [W_n, S^1] \twoheadrightarrow [E,S^1]$$ is surjective, and thus the naive cohomology group $[E,K(\mathbb{Z},1)]$ is countable. In particular, a CW approximation $X' \stackrel{\sim}{\to} X$ induces a non-surjective map on naive cohomology $[E,K(\mathbb{Z},1)] \to H^1(E;\mathbb{Z})$.

[1] Katsuya Eda and Kazuhiro Kawamura, MR 1772189 The singular homology of the Hawaiian earring, J. London Math. Soc. (2) 62 (2000), no. 1, 305--310.

Here are two examples similar to Ben's answer, followed by an example with a path-connected space.

Let $[X,Y]$ denote the naive pointed homotopy classes of pointed maps. We're comparing the (say, unreduced) singular cohomology functor $H^n(X;G)$ to the naively represented functor $[X_+, K(G,n)]$. As Chris pointed out, singular cohomology is representable in the derived sense, i.e., agrees with its value on a CW approximation of $X$.

Example 1. Let $T \subset \mathbb{R}^2$ be the topologist's sine curve $$T = \{ (x, \sin \frac{1}{x}) \mid x \in (0,1] \} \cup \{(0,0)\}.$$ Its zeroth singular cohomology is the finite product $$H^0(T;\mathbb{Z}) = \mathrm{Hom}_{\mathbb{Z}}(H_0(T;\mathbb{Z}), \mathbb{Z}) = \mathrm{Set}(\pi_0(T),\mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}.$$ Its zeroth naive cohomology is $$[T_+, K(\mathbb{Z},0)] = \mathrm{Set}(\ast,\mathbb{Z}) = \mathbb{Z}.$$ A CW approximation $S^0 \stackrel{\sim}{\to} T$ induces on zeroth naive cohomology the diagonal inclusion $\mathbb{Z} \hookrightarrow \mathbb{Z} \times \mathbb{Z}$.

Example 2. Let $X \subset \mathbb{R}$ be $$X = \{ \frac{1}{n} \mid n \in \mathbb{N} \} \cup \{0\}.$$ Let me name the points $x_n = \frac{1}{n}$ and $x_{\infty} = 0$ for convenience. The zeroth singular cohomology is the countable product $$H^0(X;\mathbb{Z}) = \mathrm{Set}(\pi_0(X),\mathbb{Z}) = \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z}.$$ The zeroth naive cohomology is the subgroup $$[X_+, K(\mathbb{Z},0)] = \{ (a_n) \in \prod_{n \in \mathbb{N} \cup \{\infty\}} \mathbb{Z} \mid (a_n) \text{ is eventually constant}\},$$ which happens to be a free abelian group on countably many generators. A CW approximation from the discrete space $\coprod_{x \in X} \{x\} = \mathbb{N} \cup \{\infty\} \stackrel{\sim}{\to} X$ induces on zeroth naive cohomology the inclusion of said subgroup.

Example 3. Let $E = \bigcup_{n \in \mathbb{N}} C_n$ be the Hawaiian earring, where $C_n \subset \mathbb{R}^2$ denotes the circle of radius $\frac{1}{n}$ centered at $(\frac{1}{n},0)$. There is a surjective map $H_1(E; \mathbb{Z}) \twoheadrightarrow \prod_{n \in \mathbb{N}} \mathbb{Z}$; see for instance Hatcher Example 1.25 or [1]. Dualization into $\mathbb{Q}$ yields an injective map $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) \hookrightarrow \mathrm{Hom}_{\mathbb{Z}}(H_1(E; \mathbb{Z}), \mathbb{Q}) = H^1(E; \mathbb{Q}),$$ which shows that the first singular cohomology $H^1(E; \mathbb{Q})$ has uncountable dimension as a vector space over $\mathbb{Q}$. Indeed, the vector space $$\mathrm{Hom}_{\mathbb{Z}}(\prod_{i \in \mathbb{N}} \mathbb{Z}, \mathbb{Q}) = \mathrm{Hom}_{\mathbb{Q}}(\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z}), \mathbb{Q})$$ has uncountable dimension since $\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{N}} \mathbb{Z})$ does. 

Now denote $W_n = \bigcup_{i=1}^n C_i \cong \bigvee_{i=1}^n S^1$ and consider the retraction $E \to W_n$ that collapses all the circles $C_i$ with $i>n$ to the basepoint. Take a CW complex $K(\mathbb{Q},1)$. Any (pointed) map $E \to K(\mathbb{Q},1)$ is (pointed) homotopic to a map which factors as $E \to W_n \to K(\mathbb{Q},1)$ for $n$ large enough. This shows that the map $$\bigoplus_{i \in \mathbb{N}} \mathbb{Q} = \operatorname{colim}_n [W_n, K(\mathbb{Q},1)] \twoheadrightarrow [E,K(\mathbb{Q},1)]$$ is surjective, and thus the naive cohomology group $[E,K(\mathbb{Q},1)]$ is countable. In particular, a CW approximation $X' \stackrel{\sim}{\to} X$ induces a non-surjective map on naive cohomology $[E,K(\mathbb{Q},1)] \to H^1(E;\mathbb{Q})$.

[1] Katsuya Eda and Kazuhiro Kawamura, MR 1772189 The singular homology of the Hawaiian earring, J. London Math. Soc. (2) 62 (2000), no. 1, 305--310.