Showing a subspace is not simply connected.

I've been reviewing previous exams and came across this question that I completely missed. I'm trying to get a better understanding of the material for my future exam. The question is as follows:

Let $K$ be a non-empty compact subset of the plane $\mathbb{R}^2$. Show that $\mathbb{R}^2 \setminus K$ is not simply connected.

I'm just not seeing how to carry out the proof. Any help would be greatly appreciated.

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$K$ is bounded, so there is some $R$ such that a circle of radius $R$ centered at the origin contains $K$. Show that this circle isn't null-homotopic. –  Qiaochu Yuan Jan 11 '13 at 3:28

$\,\emptyset\neq K\subset\Bbb R^2\,$ is closed and bounded, and thus there's some circle
$$C:=\{(x,y)\in\Bbb R^2\;\;;\;\;x^2+y^2=R^2\}\,\,\,s.t.\,\,\,K\subset\,\,\, \stackrel{\circ}C$$
But then the path $\,C\subset\Bbb R^2\setminus K\,$ cannot be nullhomotopic in $\,\Bbb R^2\setminus K\,$ ...