An open set of $ \mathbb{R}^2$ without a point is not simply connected 
An open set of $ \mathbb{R}^2$ without a point is not simply connected

I need a rigorous proof of this because I only have the intuitive idea that a loop around the point can not be deformed into a point. 
 A: Let $U$ be an open set and $x\in U$. Assume without loss of generality that $x=0$, the origin. We can define the continuous function
$$f:U\setminus\{0\}\to S^1, x\mapsto \frac x{\|x\|}.$$
Because $U$ is open, there is some $\epsilon>0$ so that $B(0,\epsilon)\subseteq U$. So take any closed curve $\gamma:[0,1]\to S^1$. You can embed this curve in $U\setminus\{0\}$ via $\gamma\mapsto \epsilon\gamma=:\gamma'$. If $U\setminus\{0\}$ is simply connected, there is a homotopy contracting $\gamma'$ to a point. Via $f$, this maps to a homotpy of $\gamma$ to a point in $S^1$. We know that $S^1$ is not simply connected, hence not every curve should be contractible. Contradiction. $\square$
A: By normalizing each nonzero vector you get a retraction of the plane minus the origin to a circle.  This is a homotopy equivalence and therefore your question is equivalent to the nonsimplyconnectedness of the circle, which you say you are familiar with.
A: Probably the easiest proof is noting that $$\int_\limits{|z-\star| = \varepsilon} \frac{dz}{z-\star} \neq 0.$$ If $U \setminus \star$ would be simply connected, the integral would be zero by Cauchy's integral theorem.
