Let $C$ be a simple closed contour lying entirely within a domain $D$. Suppose $f$ and $g$ are analytic in $D$ and that $|f(z)-g(z)|\lt |f(z)|$ for all $z$ on $C$. Then $|F(z)-1|\lt 1$, where $F(z)=g(z)/f(z)$. This inequality shows that the image $C'$ of the curve $C$ under the mapping $w=F(z)$ is a closed path.
How does "$C'$ is a closed path" follow from the inequality? I can only confirm that $C'$ lies within the open disk $|w-1|\lt 1$.