TITLE: Does a connected countable metric space exist?
QUESTION [7 upvotes]: I'm wondering if a connected countable metric space exists.
My intuition is telling me no.

For a space to be connected it must not be the union of 2 or more open
  disjoint sets.
For a set $M$ to be countable there must exist an injective function
  from $\mathbb{N} \rightarrow M$.

I know the Integers and Rationals clearly are not connected. Consider the set $\mathbb{R}$, if we eliminated a single irrational point then that would disconnect the set.
A similar problem arises if we consider $\mathbb{Q}^2$
In any dimension it seems by eliminating all the irrational numbers the set will become disconnected. And since $\mathbb{R}$ is uncountable there cannot exist a connected space that is countable.
My problem is formally proving this. Though a single Yes/No answer will suffice, I would like to know both the intuition and the proof behind this.
Thanks for any help. 
I haven't looked at cofinite topologies (which I happened to see online). I also don't see where the Metric might affect the countability of a space, if we are primarily concerned with an injective function into the set alone.

REPLY [20 votes]: Fix $x_0 \in X $. Then, the continuous(!) map
$$
\Phi: X \to \Bbb {R}, x \mapsto d (x,x_0)
$$
has an (at most) countable, connected image.
Thus, the image is a whole (nontrivial!, if $X $ has more than one point) interval, in contradiction to being countable.
EDIT: On a related note, this even show's that every connected metric space with more than one point has at least the cardinality of the continuum.