Let $X := \{0\} \cup \{\frac{1}{n} \ | \ n \in \mathbb{N}\}$ be the topological space with the subspace topology. Regard $\mathbb{N}$ as a topological space with the discrete topology. Show that there is a weak homotopy equivalence $\mathbb{N} \to X$, but that it is not a homotopy equivalence. Moreover, show that there is no weak homotopy equivalence $X \to \mathbb{N}$.
I have a faulty proof for the last part. That probably also means my proof for the first part is wrong, so I will post it here as well.
As each point in $\mathbb{N}$ is a path component, its fundamental groups $\pi_n(\mathbb{N}, n_0)$ are trivial for all $n \geq 1$, $n_0 \in \mathbb{N}$. Each element $\frac{1}{n}$ is a connected component, so also a path component. As connected components partition $X$, $0$ must be a connected component as well, hence a path component. By the same logic, $\pi_n(X, x_0)$ is trivial for all $n \geq 1$ and $x_0 \in X$. That means that any map $\mathbb{N} \to X$ is a weak homotopy equivalence. Any function is continuous, so take for example $f(0) = 0, f(n) = \frac{1}{n}$. Let $g$ be a homotopy inverse of $f$, then $f \circ g \simeq \textrm{id}_\mathbb{X}$. This homotopy defines a paths between $f(g(\frac{1}{n}))$ and $\frac{1}{n}$. That must mean that $g(\frac{1}{n}) = n$ and $g(0) = 0$. But $\frac{1}{n} \to 0$, while $g(\frac{1}{n})$ does not converge to $0$, which means that $g$ is not continuous, contradiction.
Now for the second part, suppose $f: X \to \mathbb{N}$ is a continuous map. As all homotopy groups are trivial, that means it is a weak homotopy equivalence. THere is certainly a continuous map, for example constant maps.
Best Answer
Your proof of the first part is essentially correct, but you forgot $\pi_0$. This requires to show that all path components of $X$ are singletons which is easy to prove. And now you see that you cannot take any $\phi : \mathbb N \to X$. In order that $\phi$ be a weak homotopy equivalence, it must establish a bijection between the path components of both spaces. Since these are singletons, $\phi$ must be a bijection. However, any bijection will do.
For the second part let us see what it means that a function $f : X \to \mathbb N$ is continuous.
By continuity there exists $\varepsilon > 0$ such that $\lvert f(x) - f(0) \rvert < 1$ for $\lvert x - 0 \rvert < \varepsilon$. This implies that $f(x) = f(0)$ for $\lvert x - 0 \rvert < \varepsilon$. Thus if $n > 1/\varepsilon$, then $f(1/n) = f(0)$. This shows that $f$ takes only finitely many values (which are contained in the set $\{f(0)\} \cup \{f(1/n)\mid n \le 1/\varepsilon \}$).
Hence $\pi_0(f) : \pi_0(X,x_0) \to \pi_0(\mathbb N,f(x_0))$ is never a bijection (for any $x_0$).
By the way, you can use this also to show that no bijection $\phi : \mathbb N \to X$ can be a homotopy equivalence. If it were one, then it would have a homotopy inverse $g$. This map would also be a homotopy equivalence, in particular it would give us a bijection $\pi_0(g) : \pi_0(X,0) \to \pi_0(\mathbb N,g(0))$. But this is impossible as we have seen.