Suppose that we know the Jordan curve theorem. How to prove the following theorem:
Let $K$ be a compact subset of $\mathbb{R^2}$ and the set $\mathbb{R^2} \setminus K$ has $k\in \mathbb{N}$ components. If $h:K\rightarrow h(K)\subset \mathbb{R^2}$ is a homeomorphism, then the set $\mathbb{R^2} \setminus h(K)$ has $k\in \mathbb{N}$ components.
I can see that geometrically. $K$ is homotopic to a finite number of circles glued to each other (i.e. each circle is a Jordan curve). Now we can embed this "curve" inside the unit ball $\mathbb B^3$ in $\mathbb R^3$ as follows $K\ni (x,y)\mapsto (x,y,0)\in \mathbb B^3$..
Now you can count the fundamental group $W:= \mathbb B^3\setminus K$ as follows: $W$ is homotopic (contarctable) to a space consisting of a sphere $S^2$ and finitely many (vertical infinite) cylinders live inside it.. Hence $\pi_1(W)= \bigstar_n \mathbb Z$ where $\star$ denotes the free product and $n$ is the number of components.. Clearly, any homeomorphism $h$ doesn't change $n$.