Trying to go through the proof.
Let $C_n = \sum_{k=0}^{n-1} C_k C_{n-1-k}$ with $C_0 = 1$.
$$ G(x) = \sum_{n=0}^{\infty} C_n x^n \\ G(x) = \sum_{n=0}^{\infty} (\sum_{k=0}^{n-1} C_k C_{n-1-k}) x^n \\ G(x) = 1 + \sum_{n=1}^{\infty} (\sum_{k=0}^{n-1} C_k C_{n-1-k}) x^n \\ G(x) = 1 + x\sum_{n=1}^{\infty} (\sum_{k=0}^{n-1} C_k C_{n-1-k}) x^{n-1} \\ G(x) = 1 + x\sum_{n=0}^{\infty} (\sum_{k=0}^{n} C_k C_{n-k}) x^n \\ $$ I got stuck here so I looked up the solution at this point and somehow they jump straight from the step I am on to: $G(x) = 1+x\left(\sum_{n=0}^{\infty}C_nx^n\right)^2$
How?
Since the recurrence involves sum of products taken two at a time, it is somewhat natural to consider the expansion $(G(x))^2$.
Now what is the coefficient of $x^k$ in $(G(x))^2$?
Even if you are not aware of the cauchy product rule, its is not hard to see that the coefficient of $x^k$ in $(G(x))^2$ is $C_kC_0+C_{k-1}C_1 + \cdots + C_{0}C_{k}$
So, $(G(x))^2= C_1+C_2x+C_3x^2 ....=\frac{G(x)}{x}-\frac{1}{x}$