Proving $|R(z_1)-R(z_2)|\leq |z_1-z_2|$ where $R(z_1)$ is the radius of convergence of $f$ around $z_1$. Trying to prove this:

Let $D$ be a disk and $(f,D)$ a function element. If $R(z_1)$ is the radius of convergence of the power series expansion of $f(z)$ about a point $z_1\in D$ then $$|R(z_1)-R(z_2)|\leq |z_1-z_2|$$ for $z_1,z_2\in D$.

I have a correct proof however don't understand the logic. Here's the proof

We characterize $R(z_1)$ as the radius of the largest disk, centered at $z_1$ for which $f(z)$ can be extended analytically. Hence $f(z)$ does not extend analytically to any disk containing $\{|z-z_1|\leq R(z_1)\}$ and consequently $$R(z_2)\leq R(z_1)+|z_2-z_1|$$ Interchanging $z_1$ and $z_2$ gives $$R(z_1)\leq R(z_2)+|z_2-z_1|$$ giving the required expression.

I don't understand how they jump directly to the two inequalities after the definition of $R(z_1)$. If anyone could point me in the right direction I would appreciate it.
 A: The triangle inequality gives you that
$$D_r(p) \subset D_s(q) \iff r + \lvert p-q\rvert \leqslant s.\tag{$\ast$}$$
The direction $\Leftarrow$ follows since $\lvert z-q\rvert = \lvert (z - p) + (p-q)\rvert \leqslant \lvert z-p\rvert + \lvert p-q\rvert$, so if $z\in D_r(p)$ it follows that $\lvert z-q\rvert < r + \lvert p-q\rvert$, and if the right hand side of $(\ast)$ holds, it further follows that $z \in D_s(q)$. This part is true in any metric space.
For the other direction, we note that the case $p = q$ is obvious, so we may assume $p \neq q$. Then note that for $z$ of the form $z_t = p + t\cdot (p-q)$ with $t > 0$ we have $\lvert z_t - q\rvert = \lvert z_t - p\rvert + \lvert p-q\rvert$ (this uses properties of the modulus on $\mathbb{C}$), and thus there are $z_t \in D_r(p)$ with $\lvert z_t - q\rvert$ arbitrarily close to $r + \lvert p-q\rvert$. If now $r + \lvert p-q\rvert > s$, this shows that $D_r(p) \not\subset D_s(q)$.
And thus, if we had $R(z_2) > R(z_1) + \lvert z_1 - z_2\rvert$, then $D_{R(z_2)}(z_2)$ would contain the disk $D_{R(z_1) + \varepsilon}(z_1)$ for small enough $\varepsilon > 0$. But then the Taylor series of $f$ about $z_1$ would converge on the disk $D_{R(z_1) + \varepsilon}(z_1)$, contrary to the definition of $R(z_1)$ as the radius of convergence of the Taylor series of $f$ about $z_1$.
