If $f(z)$ is analytic on a disk of radius $1$, wouldn't it be true that the taylor expansion would converge to $f(z)$ for each $z$ in this disk of radius 1?
Yes, this is true, by Cauchy's theorem on the power series expansion of an analytic function in its domain of analyticity. Since $f$ is analytic in $|z-a|<1$, $f$ equals its Taylor series there.
In that case isn't the radius of convergence $1$?
No, and that is the whole point of the example. The affirmative answer to the first question implies only that the radius of convergence of the series is at least 1. But it could be larger, and indeed it is in this case.
The problem with the statement that "the radius of convergence is the distance to the closest singularity" is that you are interpreting it as saying the distance to the closest singularity $\color{red}{\text{of $f$}}$, when you should be interpreting it as the closest singularity $\color{red}{\text{of the sum of the series}}$. In this case, the distance to the closest singularity of $f$ is 1, because the principal branch of $\sqrt{z}$ is not defined on the negative real axis, and $-1+i$ is 1 unit away from the negative real axis. But the closest singularity of the sum of the series is $0$, which is a distance $\sqrt{2}$ away.
The moral of this story is that there are really two functions, $f$ and the sum of the series, both defined in the disk $|z-a|<\sqrt{2}$. They agree on $|z-a|<1$, but they disagree on $1<|z-a|<\sqrt{2}$.
For another example, consider the behavior of
$$f(z)=\begin{cases} 0 & |\text{Re}(z)|\leq1\\e^{-1/(|z|-1)^2}&|\text{Re}(z)|>1\end{cases}$$
at $z=0$. Obviously the Taylor series there is zero, which has radius of convergence $\infty$. But $f$ only equals its Taylor series in $|z|<1$. The distance to the closest singularity of $f$ is 1, but the distance to the closest singularity of the series is $\infty$.
