Why is the Koch curve homeomorphic to $[0,1]$? Henning Makholm has provided a nice proof that the limiting curve is a continuous function from $[0,1]$ to the plane. I was curios if the function is homeomorphism. A quick search gave me many sources mentioning that $[0,1]$ and the Koch curve are indeed homeomorphic (as a proof that they have the same topological dimension), yet without showing why are they homeomorphic. What would be the simplest way to show it?

The map from the interval onto the Koch curve is a continuous bijection from a compact space to a Hausdorff one. So it's closed since closed subsets of compact spaces are compact, images of compact spaces are compact, and compact subspaces of Hausdorff spaces are closed, and thus a homeomorphism.
To check this is indeed an injection from $I$ to its image, consider the following picture of one stage in the construction. Each point on some curve $\gamma_i$ is drawn directly above the point on $\gamma_{i+1}$ mapped to by the same point in $I$. One sees that only $D$ and $F$ are closer together on $\gamma_{i+1}$ than on $\gamma_i$. But even so they will never get closer than $3^{-i}$, the length of the segment containing just $E$-at the very least no closer than half this.
To see this, observe that only the tip of the new triangle lies directly above $F$'s segment, so that we can bound the distance between the whole segment $AE$ and the next triangle to the right, (out of frame,) by $3^{-i}$. And continuing the construction we can easily bound the image of $AE$ in the limit curve away from the out-of-frame triangle by $\frac{1}{2*3^{i}}$, since in later stages no point can move further than $\frac{1}{2*3^{i}}=\sum_k 3^{-i-1-k}$. Indeed at stage $j$ no point moves further than $\frac{\sqrt{3}}{2* 3^{j}}\leq 3^{-j}$. 
So if $|\gamma_i(x)-\gamma_i(y)|>\delta$, then on the limit curve $|\gamma(x)-\gamma(y)|\geq \min(\delta,\frac{1}{2* 3^i})$, in particular, $\gamma(x)\neq \gamma(y)$ and $\gamma$ is injective.  
