If $G$ is a convex set, prove that $\mathbb{C}\cup\{\infty\}\setminus G$ is connected. Let $G$ be a convex set. I want to prove that $G$ is simply connected. This is my definition of simply connected:

A domain $G$ in $\mathbb{C}$ is said to be simply connected if $\overline{\mathbb{C}}$, its extended complement, is connected.
  A domain is a nonempty, connected, open subset of $\mathbb{C}$. 

I know this problem has been posted on this site, but I haven't found on using this definition. This definition is from Sarason's complex analysis and I do not much (or, really, any) topology to follow the other answers. 
I know this problem must be easy (the author asserts the the reader will easily verify this), but I really have no idea on how to start. 
I thought about the particular case where $G$ is the unit disk. Then if I imagine $G$ on the Riemann sphere, it's not clear to me why the complement of $G$ cannot be written as the disjoint union of open sets.

We may assume without loss of generality that $G$ is properly contained in $\mathbb C$. Now let $z$ be in $\mathbb C\setminus G$ and consider an arbitrary line in $\mathbb C$ passing through $z$. This line is the union of two rays starting at $z$. I claim that at least one of those rays is disjoint from $G$. Indeed, if both contain a point of $G$, then by convexity $z$ is contained in $G$, contradicting the choice of $z$. Now any such ray gives us a path from $z$ to $\infty $ on the Riemann sphere, hence $\mathbb C\cup \{\infty\}\setminus G$ is path-connected and therefore connected.
