Quasi circle is not contractible

Question

I'm trying to show that the quasi circle (picture below) doesn't have the homotopy type of a CW complex. I proved that all homotopy groups are zero. Now I need to show that it is not contractible to use Whitehead's theorem. I know that we can collapse the vertical interval to a point to get a circle. Maybe we can use this map to show that the quasi circle is not contractible, but I don't know how.

Thanks

Answer 1

The easiest argument is to observe that the 1st Cech cohomology of this space is nontrivial. You can see this by observing that it separates the plane in 2 components and applying Alexander duality.

Answer 2

Say we want to contract it to $(0, 0)$. Look at a point on the vertical interval. Any neighbourhood of that point contains some points of the sine-curve, but any point on the sine curve has to go all the way back through $(1, 0)$ to get to the origin. Thus a contraction is impossible to do continuously.