Family of functions with $f(0) = 0$ and $f(\mathbb{D}) \cap [1,2] = \emptyset$ is normal I have the following question

Let $B$ be the set of functions $f$, which are analytic on the unit disk $\mathbb{D}$ and satisfy both $f(0) = 0$ and $f(\mathbb{D}) \cap [1,2] = \emptyset$. Prove that $B$ is a normal family.

There are a couple parts of my answer that I am unsure about.
Consider the translated family $g(z) = f(z) - 1$ which takes values in $\mathbb{C} - [0,1]$. Since $g(\mathbb{D})$ is simply connected and nonzero, we may define single-valued analytic branches of $\sqrt{g(z)}$ in $g(\mathbb{D})$. Once we take a square root, all of the values of $\sqrt{g(z)}$ are contained in a half plane where the line separating the half planes contains the origin. Then, after a possible rotation we may assume that $\sqrt{g(\mathbb{D}})$ is contained in the left half plane. Now, I can apply techniques used in this answer $\mathcal{F} \subset \mathcal{H}(D(0,1))$ with $Re f>0$ and $f(0)=1$ is a normal family to show that the translated family (hence $B$) is a normal family.
One thing that I am unsure about is whether I can say that all the values of $\sqrt{g(z)}$ are contained in a half plane where the line seperating the half planes contains the origin. This seems true, but I am not sure. Also, I am not using the full strength of the fact $f(\mathbb{D}) \cap [1,2] =\emptyset$ as I really need only $f(\mathbb{D}) \cap \{1\} = \emptyset$.
Any comments or suggestions would be greatly appreciated.
 A: Your idea doesn't quite work, and that you didn't use the assumption that a nondegenerate interval was left out of the range should serve as a warning sign (but of course it's not in itself a proof that the argument can't work).
To see that $f(\mathbb{D}) \cap \{1\} = \varnothing$ doesn't imply normality of the family consider the functions
$$f_k(z) = 1 - e^{kz}$$
for $k \in \mathbb{N}$. We have $f_k(\mathbb{C}) \cap \{1\} = \varnothing$ for all $k$, and $f_k(0) = 1 - 1 = 0$. But $f_k(z)$ converges locally uniformly to $\infty$ in the right half-plane, and it converges locally uniformly to $1$ in the left half-plane. The sequence doesn't converge locally uniformly at any point of the imaginary axis.
The first error in your argument is the claim that $g(\mathbb{D})$ is simply connected. It needn't be, consider for example
$$g(z) = -\exp \biggl(\frac{1 + z}{1-z} - 1\biggr)\,,$$
where $g(\mathbb{D})$ is the complement (in the plane) of a small disc around $0$. The simple connectedness of $\mathbb{D}$ guarantees the existence of a holomorphic square root $\sqrt{g(z)}$, but the image of that can still be all of $\mathbb{C}\setminus \{0\}$.
But the basic idea to use the square root to get a family of holomorphic functions with image contained in one half-plane works, it just needs to be done a little differently.
Consider the Möbius transformation
$$T \colon w \mapsto 2\cdot\frac{w-1}{w-2}\,.$$
This maps the closed interval $[1,2]$ to $[-\infty, 0]$, and $T(0) = 1$.
Using this, we can consider the family
$$\tilde{B} = \Biggl\{ z \mapsto \sqrt{2\cdot \frac{f(z) - 1}{f(z) - 2}} : f \in B\Biggr\}$$
where the principal branch of the square root is used.
Now, $\tilde{B}$ is just the family considered in the linked question, hence we know it's a normal family. Then it remains to deduce normality of $B$ from that. (If $(h_k)$ is a locally uniformly convergent sequence, then $(F\circ h_k)$ is also locally uniformly convergent under mild conditions on $F$.)
