I believe that the sentence starting "That is..." is not intended to be a proof of Corollary 1.4 but rather a clarification of the statement. Let me try to restate what is being claimed:
Suppose $f(z)$ is defined and holomorphic on (at least) an open disk of radius $R > 0$ centered at $z_0 \in \mathbb{C}$. Then the radius of convergence of the Taylor series expansion of $f$ at $z_0$ is at least $R$.
This is true, and indeed it is a very standard fact in elementary complex analysis. At this point in my career it's been so long since I've studied this material (and I've never yet taught it...) that my brain has reserved a very small amount of space for remembering why these things are true. I consulted it, and it said "Use Cauchy's estimates," but then that's what it says for most results in this area! Any complex analysis text will do better.
The statement should not be interpreted to mean that if a Taylor series expansion of a function $f$ about a point $z_0$ has radius $R$, then there is necessarily at least one $z$ with $|z-z_0| = R$ such that the Taylor series diverges at $z$. Gerry Myerson's answer gives a simple example where that does not happen: $\sum_{n=1}^{\infty} \frac{z^n}{n^2}$. Nevertheless, there is a sense in which the associated function $f$ has a singularity on the boundary: $f$ cannot be analytically continued to any connected open set containing (the open unit disk and) the point $1$. This has to do with the branch cut in the dilogarithm function $\operatorname{Li}_2(z)$: see e.g. this wonderful article by Don Zagier for lots more information about $\operatorname{Li}_2(z)$.
Added: Okay, I looked at the document in question. Indeed, Corollary 1.4 as I have stated it follows immediately from Theorem 1.3, so what follows the statement is a clarification, not a proof. (And I agree that the unclarified statement of Corollary 1.4 is rather confusing...)