The Maximum Modulus Principle Applied to the Proof of Schwarz Lemma

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I am using the following statement of the Maximum Modulus Principle:

Theorem: Let $G$ be a region and let $f$ be holomorphic on $G$. Suppose $\exists~ a \in G$ such that $|f(z)| \leq |f(a)| ~ ~\forall z \in G$. Then $f$ must be constant.

Let $\triangle$ denote the open unit disk, i.e. $\triangle = \{z : |z| < 1 \}$. The following is the statement of Schwarz's Lemma which I am trying to understand the proof.

Schwarz Lemma: Suppose $f$ is holomorphic on $\triangle$, $f(0)=0$ and $|f(z)| \leq 1$ for all $z \in \triangle$. Then

(1) $|f'(0)| \leq 1 \\$

(2) $|f(z)| \leq |z|$ for all $z \in \triangle$

(3) If equality holds in statement (1) or if for some point $z_0 \in \triangle \setminus \{0\}$, we have $|f(z_0)| = |z_0|$ then $f(z) = ze^{i \theta}$ for all $z \in \triangle$ and for some $\theta \in \mathbb{R}$.

Beginning of the Proof:

Define $g(z) = f'(0)$ if $z=0$ and $g(z) = \frac{f(z)}{z}$ if $z \neq 0$. Then $g$ is holomorphic on $\triangle$. Now if $0 < |z| = r < 1,$ then by Maximum Modulus Principle, $|g(z)| \leq \frac{1}{r}$.

I have two questions regarding the statement in bold.

Question 1: Why do we need the Maximum Modulus Principle (MMP) to conclude $|g(z)| \leq \frac{1}{r}$? To me, you can conclude this without the MMP because: if $0 < |z| = r < 1,$ then $|g(z)| = \bigg| \frac{f(z)}{z} \bigg| = \frac{|f(z)|}{r} \leq \frac{1}{r}$ by our hypothesis.

Question 2: How can we even apply the MMP here? The MMP requires us to have a region G. The set of all $z$ such that $0 < |z| = r < 1$ is indeed a region but we also must have $|g(z)| \leq |g(a) |$ where $a$ is some point in the set of all $z$ such that $0 < |z| = r < 1$. But I haven't shown that such a point $a$ exists. On top of that, the MMP tells us if such an $a$ exists then $g$ must be constant which is not the conclusion of the bolded statement.

My final thoughts is that I am using the "wrong" statement of the MMP. From looking online I see there are other equivalent formulations of the MMP but for the proof of Schwarz's Lemma, I am only allowed to use the MMP stated as above, no other formulations are allowed.

Thanks!