(1) What's $\pi_1(Y)$?
(2) What's the universal covering space, $\tilde{Y}$, of $Y$?
(3) What possibilities are there for the group homomorphism $f_* : \pi_1(X) \rightarrow \pi_1(Y)$?
(4) Show there exists a lift $\tilde{f} : X \rightarrow \tilde{Y}$ with $f = p \circ \tilde{f}$, where $p : \tilde{Y} \rightarrow Y$ is the covering map.
(5) Argue $\tilde{f}$ is nullhomotopic.
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.
Best Answer
You can use a bit of covering space theory. There is a theorem that says if $f:Y\rightarrow Z$ induces the trivial map on the fundamental group, for $Y$ path connected, locally path connected then there is a lift of $f$, $\tilde{f}:Y\rightarrow \tilde{Z}$ where $\tilde{Z}$ is any covering space, i.e. $p\circ\tilde{f}=f$, where $p:\tilde{Z}\rightarrow Z$ is the covering map. See, for instance, Proposition 1.33 in Allen Hatcher's $\textit{Algebraic Topology}$. Since $S^1$ has a contractible universal cover, $\mathbb{R}$, the map $f$ factors through a contractible space and is therefore null-homotopic.