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?

You can define the Koch curve without using a sequence of functions (I dislike sequences of functions)
First define $f$ on $[0,1] \cap \Bbb Z[1/2]$ (the dyadic rationals) inductively by :
$f(0)= (0,0)$, $f(1) = (1,0)$, and for $i=0 \ldots 3$, $f((x+i)/4) = T_i(f(x))$ where $T_0,T_1,T_2,T_3$ are the $4$ affine transformations
$T_0(x,y) =(\frac x 3,\frac y 3), \\
T_1(x,y) = (\frac 13+ \frac{x-\sqrt 3 y}6,\frac{\sqrt 3 x + y}6), \\
T_2(x,y) = (\frac 12+\frac{x+\sqrt 3 y}6, \frac{\sqrt 3}2 + \frac{-x+\sqrt 3 y}6), \\
T_3(x,y) = (\frac 23 + x,\frac y3)$
If $A^0$ is the triangle with vertices $(0,0),(1,0),(\frac 12,\frac{\sqrt 3}6)$, we can see that if $A^1_i = T_i(A^0)$ then $A^1_i \subset A^0$ ; any two $A^1_i,A^1_j$ have distinct interior ; two consecutives triangles are joined by a vertex ; the diameter of $A^1_i$ is $\frac 13$ the diameter of $A^0$.
The points $f(x)$ for $x$ a dyadic rational are vertices of the images of the initial triangle by the various iterations of the $4$ transformations (with usually two possibilities).  
The facts that $T_{i+1}(0,0) = T_i(1,0)$ (two consecutive triangles are connected),and that the diameters of the triangles decrease exponentially prove that $f$ is uniformly continuous : if $|x-y| \le 1/4^n$ then $x$ and $y$ are either in the same triangle $A^n_k$ or in two consecutive triangles $A^n_k,A^n_{k+1}$, which shows that $d(f(x),f(y)) \le 2/3^n$.
The fact that the $4$ triangles $T^1_i$ are almost disjoint proves that $f$ is "uniformly injective". We can show with a few pictures that if you pick a triangle $T^n_k$ and remove its two neighbourhing triangles, the rest of the $T^n_{k'}$ triangles will be at a distance $\ge C/3^n$ for some $C > 0$. Therefore, if $|x-y| > 2/4^n$, then $d(f(x),f(y)) \ge C/3^n$.
Now we can extend the definition of $f$ on $[0;1]$ by continuity. It's easy to show that $f$ stays uniformly continuous and uniformly injective (hence it stays injective). So $f$ is a continuous bijection on its image. When you take the contrapositive of the uniform injectivity, you obtain that $f^{-1}$ is uniformly continuous, so that $f$ is a homeomorphism.
