Proof of the three-point characterization of holomorphy This post on Math Overflow is looking for the source of the following theorem:

Let $D = \{ z \in \mathbb{C} : |z| < 1 \}$ denote the open unit disk. A function $f : D \to D$ is holomorphic iff it has the following property: for any three points in $D$, there exists a holomorphic function $g : D \to D$ that agrees with $f$ at the three points.

I'm interested in the proof. One direction is trivial, and for the other direction, the post says that there is an "easy" but clever proof using only Schwarz's lemma and Montel's theorem. So I suppose the general idea is to look at the family of functions $g$ that are holomorphic and agree with $f$ at three or more points, and to find, within this family, a sequence that converges to $f$. At first I thought this might require an enumeration of a countable dense subset of $D$, but now I think maybe it's possible to show that $f$ is differentiable at $0$, and use a Möbius transformations such that $z_0 \mapsto 0$ to show differentiability at $z_0 \neq 0$.
The only other information I've been able to find online is a post on this blog, "Calculus VII", which explains why the theorem does not hold with fewer than three points nor with the codomain being all of $\mathbb{C}$.
Please do not post a full solution. This is tagged as homework.
 A: If you want a countable dense subset of $\Bbb D$ get the union $A\cup B$ where $A=(-1,1)\cap \Bbb Q$ and $B=i(-1,1)\cap \Bbb Q$. Then $A\cup B$ is dense in $\Bbb D$ because of the density of the rationals.
Now, construct a family of holomorphic functions (like the $g$). Then they are locally bounded and then from Montel's theorem there is a subsequence that converges uniformly on the compact subsets of $\Bbb D$.
"""Now use Weierstrass's convergance theorem that states:
Let $A\subset \Bbb C$ be an open and $(f_n)$ be a sequence of holomorphic functions $A\to \Bbb C$ and $f$ a function such that $f_n\to f$, uniformly on the compact subsets of $A$. Then $f$ is holomorphic in $A$.(Moreover $f'_n\to f'$  uniformly on the compact subsets of $A$-you don't need this in this proof).
The hard part is too prove that the limit of the sequence you will build, will be $f$.
You can build a triangle inside $\Bbb D$ ,by getting three arbitrary points in $\Bbb D$. Then hold steady one of these points and run the other two so to build a $Λ$. Then run one point more so to build a triangle $\triangle$. Have in mind that a triangle is compact set.(Weierstrass's theorem).
Start with the triangle and with the logic you build it,find the holomorphic functions that the sequence that will build will converge to $f$."""
I think that you should use Schwarz's lemma about fixed points. It says that If $f$ is holomorphic in the unit disk $D$ and $|f(z)|<1$ for all $z∈D$ and
 has two distinct fixed points in $D$ then $f(z)=z$ for every $z∈D$. 
Try something like $f^{-1}og$. This one has three distinct fixed points.Good luck!
