if $f: \Omega \to \mathbb{C} \setminus \{0\}$ is holomorphic and $\Omega$ is simply connected, show there is a holomorphic function $g^2 = f$.

It is easier to prove that logarithms exist, that is, you can find some $h$ with $\exp h = f$. Once you get such $h$, it is clear that $\exp\left(\frac 1 2 h\right)$ will be a square root of $f$.

To find $h$, notice that the equation $\exp h = f$ implies that $\exp h\cdot h' = f'$ which tells you that $h' = f'/f$. The last equation makes sense because $f$ is non-vanishing (per your hypothesis).

Thus, finding logarithms is equivalent to proving that $f'/f$ has a primitive in your simply connected domain $\Omega$. Of course, this is true, since you can pick a point $z_0\in \Omega$ and set

$$h(z) = \int_{z_0}^z \frac{f'(w)}{f(w)}dw.$$

where we integrate along any path $\gamma: z_0\to z$. This is well defined because $\Omega$ is simply connected and $f'/f$ is analytic. Checking that $h'(z) = f'(z)/ f(z)$ is a direct computation.