Continuous function from a countable set to $\mathbb{Q}$. Let $N \subset \mathbb{R}$ be countable. I'd like a continuous map 
$$f:N \to \mathbb{Q}.$$  
Is it enough to fix $n \in N$ and exploit a homeomorphism contraction with $\delta > \epsilon$, $$B_\delta (n) \to B_\epsilon (f(n))?$$
Clearly this map is a homeomorphism, though I don't know how to write it down explicitly.  
I ask because I was providing an alternate proof to show that countable sets are disconnected in $\mathbb{R}$. Ergo, it suffices to show that there exists a continuous mapping from any countable set into $\mathbb{Q}$ which is totally disconnected.
 A: I don't understand the proof strategy you're proposing, but here's how I would construct a continuous injection $f:N\to\mathbb{Q}$.  Let $P=N\cup\mathbb{Q}$; then $P$ is a countable dense subset of $\mathbb{R}$.  By a well-known back-and-forth argument (see here for instance), any two countable dense subsets of $\mathbb{R}$ are order-isomorphic, and you can show that such an order-isomorphism is also a homeomorphism with respect to the subspace topologies inherited from $\mathbb{R}$.  In particular, there is a homeomorphism $P\to\mathbb{Q}$.  Composing this with the inclusion map $N\to P$ then gives a continuous injection $N\to\mathbb{Q}$.
A: If $\mathbb N$ has the discrete topology, any map $f\colon\mathbb N\to X$, where $X$ is a topological space, will be continuous.
A: Your main task is much easier to prove though. Consider a countable subset $N\subset \mathbb{R}$ and some $x<y \in N$. Then there exists $z\in (x,y)$ such that $z\notin N$ because $N$ is countable. Now you can easily split $N$ into two relatively open subsets by intersecting it with $(-\infty,z)$ and $(z,\infty)$.  If you want to stick to your previous tactic, try finding a surjection (only then is the image disconnected) into some simpler set (it doesn't have to be totally disconnected).
