Combine the hint above with this case when $\alpha = i$:
Consider the Cayley transform $W(z) = \frac{z-i}{z+i}$ which maps the upper half plane conformally onto the unit disk. Now, $W \circ \phi (0) = W(\alpha) = 0$. Hence, $W \circ \phi$ is a conformal map of the unit disk to the unit disk that fixes $0$, so Schwarz applies. Thus $|(W \circ \phi)'(0)| \leq 1$. By the chain rule, $(W \circ \phi)' (0)= W'(\phi(0)) \phi'(0) = W'(i) \phi'(0)$. Now $W'(z) = \frac{2i}{(z+i)^2}$, so $W'(i) = -i/2$. Thus
$$|\phi′(0)| = | \frac{(W \circ \phi )' (0) }{ W'(i)} | \leq 2.$$
In this proof I'll show you how to find a conformal map from S to the upper half plane. Then you can take the inverse and compose it with a Möbius transform.
Consider the function $g(z)=\dfrac{1}{z+1}$, which is a Möbius transform, therefore it maps circles/lines to circles/lines. Since it has a pole in -1, we conclude that it maps the interval $[-1,1]$ , as well as the upper circle ( it is the set $\partial S\setminus (-1,1)$) to two lines, which intersect orthogonally at the point $g(1)=\frac{1}{2}$.
By taking the image of different points of S you can see that the image of S via g is the 4th quadrant, translated by 1/2 to the right, i.e. $\{(x,y)\in\mathbb R^2:x>\frac{1}{2},y<0\}$.
As a result, $g(z)-\frac{1}{2}$ maps S to the 4th quadrant.
Continuing, we rotate by $\pi/2$, so $e^{i\frac{\pi}{2}}(g(z)-\frac{1}{2})=i(g(z)-\frac{1}{2})$ maps S to the 1st quadrant.
Finally, by squaring, we can map the 1st quadrant to the upper half plane, in other words:
$$(i(g(z)-\frac{1}{2}))^2=\dfrac{-(z-1)^2}{4(z+1)^2}$$
maps S to the upper half plane.
Note: now compose with $\dfrac{z-i}{z+i}$ which maps the upper half plane to the unit disk , to get a conformal map from S to D. What you need is the inverse of this map.
Best Answer
Geometrically $\{z\in \Bbb C | \ \ |z+1|=|z-1|\}$ is locus of all $z$ which is equidistant fron $1$ and $-1$, which is nothing but imaginary axis in complex plane.
Whereas $A=\{z\in \Bbb C | \ \ |z+1| \geq |z-1|\}$ is right half plane. $\tag{1}$
Hence the map $f:A \to \Bbb D$ defined as $$z \mapsto \frac{z-1}{z+1}$$ and from $(1)$ we can infer that $f$ is well-defined and is bijective map.
$g:A \to \Bbb H$ defined as $z\mapsto iz$ rotates the right half plane to upper half plane.
Hence the desired map $g\circ f^{-1}: \Bbb D \to \Bbb H$.