TITLE: Countable path-connected Hausdorff space
QUESTION [7 upvotes]: Is there a topology $\tau$ on $\omega$ such that $(\omega,\tau)$ is Hausdorff and path-connected?

REPLY [14 votes]: I realise this is quite an old question, but here is an alternative to the already existing elegant answers, which shows that Hausdorff can be replaced by $T_1$:

Proposition. No countable $T_1$ space containing at least two points is the continuous image of an interval.

Proof. Suppose $X$ is a $T_1$ topological space (so points are closed), $I:=[0,1]$, and $f\colon I\to X$ is continuous. Then
$$ \{f^{-1}(x)\colon x\in X\}$$
is a partition of $I$ into closed, hence compact, subsets. But $I$ (more generally, any non-trivial continuum) is not the countable disjoint union of compact proper subsets. So $X$ is uncountable. $\square$
Observe that the proof, in contrast to Joel's, does not show that the cardinality of $X$ is that of the continuum. On the other hand, there are non-Hausdorff compact $T_1$ spaces, so his proof does not apply to $T_1$ spaces. Likewise for Todd's argument, as there are $T_1$ spaces that are path-connected, but not arcwise connected. 
On the other hand, Sierpiński space (consisting of two points, only one of which is closed) is both $T_0$ and path-connected. So $T_1$ cannot be replaced by $T_0$.
EDIT. It seems this argument is also here: http://topospaces.subwiki.org/wiki/Path-connected_and_T1_with_at_least_two_points_implies_uncountable