If $|f(z)| < 1$ then $|f(z)| < |z|$ Exercise Suppose $f(z)$ is analytic on $|z| <1$ and $|f(z)| < 1$ for all $|z| < 1$; also $f(0) = 0$. Show that $|f(z)| \leq |z|$ and that $|f'(0)| = 1$.
Proof Attempt.
First, I attempt to show that $|f(z)| \leq |z|$.
Case 1: |z| =0: If $z = 0$, then $f(0) = 0$. Notice that $|f(0)| = 0 \leq |0| = 0$.
Case 2: 0<|z|<1: ???
It's hard to imagine how Case 2 holds, given the below counterexample
Consider  for $|z| < 1$ the function $f(z) = z + \frac{|z-1|}{2}$
. Then for all $ 0<|z| < 1$, we have $|f(z)| < 1$ but $|f(z)| > |z|$. Perhaps there is something I am misunderstanding here?? Any help or tips are greatly appreciated.

Use the Riemann mapping on removable singularities to show that $f(z)/z$ can be extended to an analytic function, and use the max principle on the disk centered at 0 of radius $0<r<1$. You will get a bound that holds for every $0<r<1$, and then take a suitable limit to get your bound.
Now, your example is not a holomorphic function. One quick way to see this: if it were, then $f(z)-z$ would be a holomorphic real valued function defined on a open connected set. What does the open mapping theorem tell you? Of course, you can still use the CR equations to show that.
