Is the boundary of every compact connected subset ( with more than one point ) of $\mathbb R^2$ is Homeomorphic with $S^1$ ? Is it true that the boundary of every compact connected subset ( with more than one point ) of $\mathbb R^2$ is Homeomorphic with $S^1$ ? I was thinking that union of two closed balls touching tangentially might be a counterexample , but I am not sure . Please help. Thanks in advance  
 A: No, this is false. Take an annulus $\{(x, y) \in \Bbb R^2 : 1 \leq x^2 + y^2 \leq 2\}$. This is obviously connected and compact, but the boundary is disjoint union of two copies of $S^1$, not homeomorphic to $S^1$ as it's not connected.
A: Your counterexample is correct. You can add any finite number of tangent balls.
Another counterexample is the union of two segments with an intersection point. Or any path $\gamma:[0,1]\to\Bbb R^2$ that intersects itself.
A: Two interesting examples: $\bullet$1. The set $C=A\cup B$ where $A=\{0\}\times [-1,1]$ and $B=\{(x, \sin \frac {1}{x}) : 0<x\leq 1\}.$ Since $C$ is closed but has empty interior, we have $\partial C=C.$ Now $C$ is connected but not path-connected : There is no continuous $f:[0,1]\to C$ with $f(0)=(0,0)$ and $f(1)=(1,\sin 1).$...$\bullet$2.  The regular-closed set (equal to the closure of its interior) $D =\{[-1,0]\times [-2,2]\}\cup \{(x.y): x\in (0,1]\land |y-\sin \frac {1}{x}|\leq 1\}.$ The  boundary of $D$ is  connected but not path-connected. 
