Let $f(z)$ be an entire function such that $|f(z)|\geq |z|^a$ in $\overline{D}$ for some $a\in\mathbb{R}$ with $0<a<1$. Prove that $|f(0)|\geq 1$
I am thinking of the function $g(z) = \frac{z^a}{f(z)}$. If $f(0)\neq 0$, then this function is analytic in $D$, and we have $g(0)=0$, $|g(z)|\leq 1$ in $D$. By Schwarz Lemma, $|g(z)|\leq |z|$ in $D$, and hence $\frac{1}{|f(z)|}\leq |z|^{1-a}< 1$ when $0<|z|<1$. Hence we have $|f(z)| > 1$ for $z\in D\backslash\{0\}$. By continuity, we have $|f(0)|\geq 1$.
However, I don't know how to handle the case for $f(0) = 0$ (The only possible zero for $f$ in $D$ is $0$). As the question indicates, this should lead to a contradiction.
The condition $|f(z)|\geq |z|^a$ implies that $|f(z)|\geq 1$ for $|z| = 1$. If $f$ has no zeros in the unit disc then the maximum modulus principle can be applied to $1/f$, so that $$ \frac{1}{|f(0)|} \le \max \{ 1/|f(z)| : |z| = 1 \} \le 1. $$
As you said the only possible zero of $f$ is at $z=0$. But then $f(z) \sim cz^n$ for $z \to 0$ with some $c \ne 0$ and a positive integer $n$, so that $$ \frac{|f(z)|}{|z|^a} \sim c |z|^{n-a} \to 0 $$ for $z \to 0$, in contradiction to the assumption.
Your approach considering $g(z) = z^a/f(z)$ does not work because $z^a$ cannot be defined as a holomorphic function for $0 < a < 1$.