Of course it's valid for real coefficients because real number is a subset of complex number and the proof doesn't make any assumption whether the coefficient is complex or real.
For a complex $d$
$$
\begin{aligned}
|A(z)|^2 &= A(z)A^*(z) = \frac{1-d^*z}{z-d} \frac{1-dz^*}{z^*-d^*}\\
&=\frac{1-(d^*z+dz^*)+|d|^2|z|^2}{|z|^2 - (d^*z+dz^*) + |d|^2}
\end{aligned}
$$
We know that the numerator and denominator of the above equation are both non-negative. To compare it with 1, subtract denominator with numerator
$$
1 + |d|^2|z|^2 - |z|^2-|d|^2 = (1-|d|^2)+ (|d|^2-1)|z|^2 = (1-|d|^2)(1-|z|^2)
$$
If $d$ is real, then
$$
\begin{aligned}
|A(z)|^2 &= A(z)A^*(z) = \frac{1-dz}{z-d} \frac{1-dz^*}{z^*-d}\\
&=\frac{1-d(z+z^*)+d^2|z|^2}{|z|^2 - d(z+z^*) + d^2}
\end{aligned}
$$
Numerator minus denominator equals to
$$
1 + d^2|z|^2 - |z|^2-d^2 = (1-d^2)+ (d^2-1)|z|^2 = (1-d^2)(1-|z|^2)
$$
For a stable all-pass filter we have $|d|<1$, so the sign of subtraction is determined by $1-|z|^2$, i.e., if
- $|z| > 1$, $|A(z)| < 1$
- $|z| = 1$, $|A(z)| = 1$ (that's an all-pass!)
- $|z| < 1$, $|A(z)| > 1$
It seems there is something wrong with your question, it is easy to see that $|A(z)|=-1/d\neq 0$ when $z=0$.