Here is a proof that uses only Schwarz' lemma and the triangle inequality.
Set $a=f(0)$ and
$$\phi_a(z)=\frac{z-a}{1-\overline{a}z}.$$
Then $\phi_a\circ f$ maps the unit disc to the unit disc and fixes zero. Thus, by Schwarz' lemma,
$$|\phi_a(f(z))|=\left|\frac{f(z)-a}{1-\overline{a}f(z)}\right|\leq|z|,$$
and so
\begin{equation}|f(z)-a|\leq|z||1-\overline{a}f(z)|\leq|z|+|z||a||f(z)|.\tag{1}\end{equation}
Applying the triangle inequality, we arrive at
$$|f(z)|\leq|z|+|z||a||f(z)|+|a|.$$
Then
$$|f(z)|-|z||a||f(z)|\leq|z|+|a|$$
$$|f(z)|\leq\frac{|z|+|a|}{1-|a||z|}=\frac{|f(0)|+|z|}{1-|f(a)||z|},$$
Where the final equality uses the fact that $f(0)=a$. This is the second desire inequality.
To obtain the first inequality, we begin with
$$|a|=|a-f(z)+f(z)|\leq |a-f(z)|+|f(z)|\leq |z|+|z||a||f(z)|+|f(z)|,$$
where the last inequality follows from (1). Then
$$|a|-|z|\leq |z||a||f(z)|+|f(z)|$$
$$\frac{|a|-|z|}{|z||a|+1}=\frac{|f(0)|-|z|}{1+|f(0)||z|}\leq |f(z)|.$$
Instead of introducing auxiliary Möbius transformations every time, one could prove the general Schwarz-Pick lemma
$$\left|\frac{f(a)-f(b)}{1-f(a)\overline{f(b)}}\right|\le \left|\frac{a-b}{1-a\overline{b}}\right|,\quad a,b\in\mathbb D \tag1$$
valid for any holomorphic map $f:\mathbb D\to\mathbb D$. The proof is, of course, an application of Schwarz lemma to the composition of $f$ with Möbius transformations.
You can think of the quantity $d(a,b)=\left|\frac{a-b}{1-a\overline{b}}\right|$ as a kind of distance adapted to the geometry of the unit disk. Then (1) simply says that this distance does not increase under $f$. In your case,
$$d(f(4/5),f(1/2))\le d(4/5,1/2)=1/2$$
and since $f(1/2)=0$, the left side simplifies to $d(f(4/5),0)=|f(4/5)|$.
If, instead of $f(1/2)=0$, you were told that $f(1/2)=1/3$, extra work would be needed to decipher the inequality
$$\left|\frac{f(4/5)-1/3}{1-f(4/5)/3}\right|\le \frac12$$
Geometric interpretation helps here: the inequality describes a non-Euclidean disk with non-Euclidean center $1/3$. Its farthest point from the origin $0$ will be found on the radius through $1/3$. This reduces the task to solving the equation
$$\frac{x-1/3}{1-x/3}=\frac12$$
for real $x$. The solution is $5/7$. Thus, $|f(4/5)|\le 5/7$ in this version of the problem.
The quantity $\rho(a,b)=\tanh^{-1}d(a,b)$ is known as the Poincaré metric on the unit disk, and has a natural generalization to other domains in $\mathbb C$. The Schwarz-Pick lemma generalizes accordingly: a holomorphic map $f:\Omega_1\to\Omega_2$ is a contraction with respect to the hyperbolic metrics of $\Omega_1$ and $\Omega_2$.
The hyperbolic metric and geometric function theory by Beardon and Minda is an introductory article which explains all this and more; probably more than you need to know for a qualifying exam.
Best Answer
It helps to be familiar with the Möbius transformations that map the disk onto itself. One of these is $g(z)=\frac{\frac{2}{3}-z}{1-\frac{2}{3}z}$. Then $g$ is holomorphic on the disk and maps the disk bijectively to itself, and it swaps $\frac{2}{3}$ and $0$. Applying Schwarz's lemma to $g\circ f$ yields $|f'(0)|\leq \left|g'\left(\frac{2}{3}\right)\right|^{-1}$, and you can check that this upper bound is attained when $f=g$.
In general, if $f$ maps the disk to itself and $f(0)=a\neq 0$, you can compose with $\phi_a(z)=\frac{a-z}{1-\overline{a}z}$ to get a map $\phi_a\circ f$ from the disk to itself that sends $0$ to $0$, so that Schwarz's lemma can be applied. Notice that $\phi_a$ swaps $0$ and $a$, which is part of why $\phi_{2/3}$ is so helpful in this problem. More generally, if $f(b)=a$, then Schwarz's lemma can be applied to $\phi_a\circ f\circ \phi_{b}$.