Prove that the open ball $B(0,1) \in \mathbb{C}$ is connected. Using the theorem, An open set $G \subset \mathbb{C}$ is connected iff for any two points $a,b$ in $G$ there is a polygon from $a$ to $b$ lying entirely in $G$.
I know an open ball is indeed an open set. I'm having trouble formulating the proof that if I have two points $a$ and $b$. I can form a polygon from $0$ to $a$ and another polygon from $0$ to $b$. Since both polygons share an endpoint, $0$, then they form one polygon. 
How would I go about starting this proof?

Why don't you simply join $a$ and $b$ with a segment?  
Let $a,b\in B(0,1)$ the $t\to ta+(1-t)b$ for $t\in [0,1]$ is a parametrization of the segment that joins $a$ and $b$. All the points along this segment are inside $B(0,1)$ because for all $t\in [0,1]$,
$$|ta+(1-t)b|\leq  t|a|+(1-t)|b|< t\cdot 1+(1-t)\cdot 1=1.$$
