If $U$ is an open subset of euclidean plane $R^2$, and $f: U \rightarrow R^2$ is continuous and injective, then $f(U)$ is open in $S^2$ and the inverse function is continuous.
This is Theorem 62.3 in Topology, Munkres. According to this book,
Step 1. We show that if $B$ is any closed ball in $R^2$ contained in $U$, then $f(B)$ does not separate $S^2$.
Let $a$ and $b$ be two points of $S^2-f(B)$. Because the identity map $i: B \rightarrow B$ is nulhomotopic, the map $h:B \rightarrow S^2 - a - b$ obtained by restricting $f$ is nulhomotopic.
But, I coudln't understand how restriction of $f$ be nulhomotipic.