Proof $f'(z)=0$ implies $f(z)$ locally constant. I am currently reading "Theory of Complex Functions" by Remmert and I encountered this theorem:

Let $f:D\rightarrow\mathbb{C}$ be a complex holomorphic function. If $f'(z)=0, \forall z\in D$, then $f$ is locally constant in $D$.

Here is part of the proof:

Consider any open ball $B=B_{r}(b)\subset D$ and any $z\in B$. Let $L$ denote the line segment from $b$ to $z$ and let $\epsilon>0$ be given. For each $c\in L$, there is a disc $B_{\delta}(c)\subset D$,$\delta=\delta(c)>0$, such that $|f(w)-f(c)|<\epsilon|w-c|,\forall w\in B_{\delta}(c)$.
Because finitely many discs $B_{\delta}(c)$ suffice to cover the line $L$, there is a succession of points $z_{0}=b,z_{1},\cdots,z_{n}=z$ on $L$ such that $|f(z_{i})-f(z_{i-1})|<\epsilon|z_{i}-z_{i-1}|, 1\leq i\leq n$." $\cdots$

I understand why finitely many discs cover $L$ since it is compact. My question is how do you prove that there is a succession of points $z_{0},\cdots,z_{n}$ such that $|f(z_{i})-f(z_{i-1})|<\epsilon|z_{i}-z_{i-1}|$? The statements seems intuitively obvious but I can't provide the rigorous argument.
 A: Hopefully this is good enough.
Let $B_{\delta(c_1)}(c_1),B_{\delta(c_2)}(c_2),\ldots B_{\delta(c_n)}(c_n)$ be the finite cover of $L$ by discs. Then for each $c_i$, $|f(w)-f(c_i)|<\varepsilon |w-c_i|$ when $w\in B_{\delta(c_i)}(c_i)$. (This is terrible notation) for each pair $c_i,c_{i+1}$, we can find a $c_{i+1/2}$ which belongs to $B_{\delta(c_i)}(c_i)\cap B_{\delta(c_{i+1})}(c_{i+1})$. Set $z_1=c_1,$ $z_2=c_{1+1/2}$ $z_3=c_{2}$, etc.
Now given any $z_i$ and $z_{i-1}$, one of these points is the center of a disc and the other is contained in it. WLOG suppose it is $z_i$ that belongs to the center of a disc. Then
$$
|f(z_i)-f(z_{i-1})|<\varepsilon |z_i-z_{i-1}|
$$
and this set $z_1,\ldots z_m$ fits the bill.
I'm sure you've drawn a picture of this situation already. Just take the centers of the discs along with a point of intersection for each pair of disc which intersect.
It seems like a "connectedness argument" from topology would work for this as well, and could be easier depending on how much you know. For this argument you would fix a point $z$ in $D$ and find a $r>0$ so that $B_{r}(z)\subset D$. Suppose that $f(z)=p$ for some $p\in\mathbb{C}$. To finish the proof you must just show that the the set $\{w\in B_{r}(z)\,:\,f(w)=p\}$ is both open in closed in the subspace topology of $B_{r}(z)$: $B_{r}(z)$ is obviously connected, and the only open and closed subsets of a connected space are the empty set and $B_{r}(z)$ itself, in which case it would have to be $B_{r}(z)$ (it is nonempty since $z$ is a member). The set $\{w\in B_{r}(z)\,:\,f(w)=c\}$ is the inverse image of a closed set under a continuous function, so it is closed, and so the only hard part is to show that this set is open. Depending on how much you know this might not even be hard.
A: Here is a complete, short, and rigorous construction of the sequence $z_0,...,z_n,$ obtained by applying the Lebesgue number lemma, a standard tool for these kinds of arguments.
By continuity it follows that for each $c \in L$ there exists an open ball $B(c)$ centered on $c$ with the property that for any $w,z \in B(c)$ we have $|f(w)-f(z)| < \epsilon |w-z|.$ (My choice of $B(c)$ is more restrictive than in the text you quote). 
Since $L$ is a compact metric space, by the Lebesgue number lemma there exists $\lambda>0$ such that for every subset $A \subset L$, if the diameter of $A$ is less than $\lambda$ then there exists $c$ such that $A \subset B(c).$
Choose $n > \frac{\text{Length}(L)}{\lambda}$. Let $b=z_0,z_1,...,z_n=z$ be the sequence of points on $L$ that subdivides it into subintervals of length equal to $\frac{\text{Length}(L)}{n} < \lambda$. 
For each $i=1,...,n$ the set $\{z_{i-1},z_i\}$ has diameter $<\lambda$ and so $|f(z_{i-1}) - f(z_i)| < \epsilon |z_{i-1}-z_i|$.
