In the above proof, next to the area with the blue line, Apostol assumes that the arbitrary point $x$ that he is trying to prove that $V(x)-V(c)<\epsilon$ for is less than $x_1$ thus he can say "Adding more points to $P$ can only increase the sum". But if say $x>x_1$ we cannot use the inequality $$V_f(c,b)-\frac{\epsilon}{2}<\frac{\epsilon}{2}+\sum^n_{k=2}|\triangle f_k|\le\frac{\epsilon}{2}+V_f(x_1,b).$$ Can someone clarify whether Apostol is wrong or if there is something I am missing here.
On
$x_1$ is essentially an arbitrary point. If we wanted to test another point $x \in B_\delta(c)$ with $x>x_1$, then add it to the partition and run the same argument with $x$ replacing $x_1$.
Edit: To address the question of testing another point in the partition (not $x_1$), just add enough points so that you have $p \in B_\delta(c)$. Nothing in the argument hinges on $x_1$ being the first point in the partition after $c$.
It's minor issue here which can be fixed easily. Let $\epsilon>0$ and $\delta'>0$ be such that $$|f(x) - f(c) |<\frac{\epsilon} {2}$$ whenever $|x-c|<\delta '$. And let $$P=\{c=x_0,x_1,x_2,\dots,x_n=b\}$$ be a partition of $[c, b] $ such that $$V_f(c, b) - \frac{\epsilon} {2}<\sum_{i=1}^{n}|f(x_i)-f(x_{i-1})|$$ Let $\delta=\min(\delta', x_1-x_0)$ and consider point $x$ such that $0<x-c<\delta$. Also let $$P'=P\cup\{x\} =\{c=x'_0,x=x'_1,x'_2,\dots,x'_m=b\}$$ and continue the argument of Apostol with above partition.