Here's my attempt at an explanation.
Since the closure of $\Omega$ is compact, so is its image under $f$. By the Open Mapping Theorem, $f(\Omega)$ is open, so the boundary of that image must be $f(\partial \Omega)$, the image of the boundary of $\Omega$. We know that $f(\partial \Omega)$ is in the unit circle, so $f(\Omega)$ must be the unit disk.
Suppose $\gamma_j$ is one of the interior boundary curves of $\Omega$, oriented positively (i.e. counterclockwise). Thus as you travel around $\gamma_j$ in the forward direction, your right hand is in $\Omega$ and your left hand is in a "hole" in $\Omega$. Now suppose your friend travels on $f(\gamma_j)$ (which is on the unit circle) as you go around $\gamma_j$, so the friend is at $f(z)$ when you are at $z$. By the conformal property of analytic functions, your friend's right hand must also be in $f(\Omega)$. Thus as you go around $\gamma_j$ counterclockwise, your friend is going around the unit circle clockwise. When you get back to your starting point, your friend must also get back to his starting point, having gone at least once clockwise (i.e. in the "negative" direction) around the unit circle.
On the other hand (so to speak), if $\gamma_1$ is the outer boundary curve of $\Omega$, oriented counterclockwise, as you travel around $\gamma_1$ counterclockwise your left hand and your friend's are in $\Omega$ and $f(\Omega)$ respectively, so your friend is traveling counterclockwise around the unit circle.