Zeros of a holomorphic function

This is true and it is a consequence of the Argument Principle; the Maximum Modulus Principle is not precise enough to prove it. Here is the proof for $m=0$ and it generalizes: We know that $$n(w)=\frac{1}{2\pi i}\int_{\partial \Omega}\frac{f'(z)}{f(z)-w}\, dz$$ gives the number of times $f$ equals $w$ in $\Omega$ and is continuous on $D=\{|w|<1\}$. Since $f$ is not constant, $n(w)\ge 1$ for some $w\in D$. By connectedness and continuity, we also have $n(0)\ge 1$. Therefore $f$ has at least one zero in $D$.

In the general case, let $\partial\Omega=\gamma_1\cup\cdots\cup \gamma_{m+1}$ with orientation $\gamma_1-\cdots- \gamma_{m+1}$ (so $\gamma_1$ is also the boundary of the unbounded compoent of ${\mathbb C}\backslash\Omega$). Using the Open Mapping Theorem, one can deduce that each $f(\gamma_k)$ traverses the unit circle at least once, with $f(\gamma_1)$ in the positive direction and $f(\gamma_k)$ in the negative direction for $k\ge 2$. So each point on the unit circle is assumed by $f$ at least $m+1$ times on $\partial\Omega$. Use this fact and a point $w$ close to the unit circle to see that there are at least $m+1$ points in $\Omega$ that has image $w$. So $n(w)\ge m+1$. Now use the same argument as the $m=0$ case to conclude that $n(0)\ge m+1$. (Or one can just use an analytic continuation of $f$ and do away with $w$ close to the unit circle.)

You should say "at least $m+1$ zeros counting multiplicities", for clarity. Without multiplicities, the statement would not be correct for $m>0$ (since the map must have critical points).

Steve's answer can be phrased concisely if you know about proper maps. A continuous function between two nonempty open subsets $U$ and $V$ of the complex plane is proper if every preimage of a compact subset of $V$ is compact. This is equivalent to saying that $f(z)\to\partial V$ as $z\to \partial U$.

A proper holomorphic map has a degree; that is, there is a number $m$ such that $f^{-1}(v)$ has $m$ elements (counting multiplicity) for every $v\in V$. Indeed, the set $f^{-1}(v)$ is compact and discrete, and hence finite, for every $v$, and using the argument principle (similarly as in Steve's answer), you see that the number of elements of this set depends continuously on $v$; so this number is constant.

In your example, every point on the unit circle has $m+1$ preimages on the boundary, which implies that the degree of $f$ is at least $m+1$.

(Note that it is not necessary to assume that the boundary consists of analytic curves.)

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.