Proving that the Calkin-Wilf tree enumerates the rationals.

Let's use a binary encoding for our way from the top of the tree to a particular fraction. We encode the starting position and each move to the right as $\color{red}1$ while a move to the left is $\color{blue}0$. (This leads to the familiar binary representation of the Calkin-Wilf sequence.)

Now, if we move to the right, we move from $\frac ab$ to $\frac{a+b}b=\frac ab+1$. If we make $n$ consecutive steps to the right, we move from $\frac ab$ to $\frac ab+\color{red}n$, i.e. we effectively add $n$. If we move to the left, we move from $\frac ab$ to $\frac a{a+b}$, which is the same as $\frac1{1+\frac ba}$. $n$ steps to the left means moving from $\frac ab$ to $\frac1{\color{blue}n+\frac ba}$.

Now, hopefully this already rings a bell: this will lead to continued fractions! Let's take as an example $\frac 37$ which, as a continued fraction, looks like so: $$ 0 + \frac1{2+\frac13} = [0;2,3] $$ We can view this representation, reading it backwards, as a means to navigate the tree from the top. To reach $3$, you need three $\color{red}1$s. (The first one is for the starting point $\frac11$, the other two are to go from there to $\frac11+\color{red}2$.) To go from $3$ to $\frac1{\color{blue}2+\frac13}$, you need two $\color{blue}0$s. At that point, you're already done; the last zero means you don't need to take any more "$\color{red}1$ turns".

Here's another example. $\frac{19}{11}$ is $$ 1 + \frac1{1+\frac1{2+\frac1{1+\frac12}}} = [1;1,2,1,2] $$ Reading $[1;1,2,1,2]$ backwards translates to two $\color{red}1$s ($\frac{\color{red}2}1$), one $\color{blue}0$ ($\frac1{\color{blue}1+\frac12}=\frac23$), two $\color{red}1$s ($\frac23 + \color{red}2=\frac83$), one $\color{blue}0$ ($\frac1{\color{blue}1+\frac38}=\frac8{11}$), and finally one $\color{red}1$ ($\frac8{11}+\color{red}1=\frac{19}{11}$).

So, to reach any rational number, compute its continued fraction and read it backwards which'll give you a way to navigate the Calkin-Wilf tree and find the number, thereby re-creating the number, step by step, from the continued fraction. This also proves that every positive rational number actually occurs in the tree.

There's one small catch which is that you obviously must start with $\color{red}1$s and end with $\color{red}1$s (possibly, as in the first example, with zero $\color{red}1$s). Which means the continued fraction needs to consists of an odd number of pieces. This is not a problem, but rather a good thing. There are exactly two continued fraction representations for each rational number, and exactly one of them is the one we need. (For example, $\frac38$ can be written as $[0;2,1,2]$, which won't work, but also as $[0;2,1,1,1]$, which is what we need.) This proves uniqueness in the Calkin-Wilf tree.