The closed topologist's sine curve is the set
$\{(x,y)\in\mathbb{R}^2 : x=0, -1\le y\le 1\}\cup\{(x,y)\in\mathbb{R}^2 : y=\sin\frac{1}{x}, 0
Clearly it is not possible to continuously map the closed unit interval onto this set, as it fails to be path connected. But what about the disjoint union of two intervals?
I would guess not, I would hope the topologist's sine curve is more badly behaved, but I can't see how to show it's not possible. Is the continuous image of a locally (path) connected space, necessarily locally (path) connected under some reasonable assumptions?