Is this property true? Suppose $f$ is a continuous function on the compact set $[0,1]$. Is true that given $\epsilon>0$, there exist $\delta>0$ such that for any partition $P=\{x=x_0<x_1,...,<x_{i-1}<y=x_i\}$ of the interval $[x,y]$, with $|x-y|<\delta$, we have that $$\sum_{n=1}^i|f(x_n)-f(x_{n-1})|<\epsilon$$
 A: Consider the function
$$f(x)=\begin{cases}
0,&\text{if }x=0\\
x\sin\frac1x,&\text{if }0<x\le 1\;;
\end{cases}$$
clearly $f$ is continuous, but you can make
$$\sum_{n=1}^i|f(x_n)-f(x_{n-1})|$$
as large as you like when $x=0$, even if $y$ is very small: if you choose the right partition, you’re essentially adding up terms of the harmonic series.
A: Not in general.  Every continuous function that is not of bounded variation provides a counterexample.  
If $f$ is not of bounded variation, then for all $\delta>0$, there exists an interval $[x,y]\subset [0,1]$ with $|x-y|<\delta$ and there exists a partition $P=\{x=x_0<x_1,...,<x_{i-1}<y=x_i\}$ such that $\sum_{n=1}^i|f(x_n)-f(x_{n-1})|>1$.  This can be seen by breaking up $[0,1]$ into finitely many intervals of length less than $\delta$, because $f$ cannot have bounded variation on all of these without having bounded variation on the entire interval.
This property is implied by absolute continuity, but not conversely.  E.g., every monotone continuous function satisfies this property, but need not be absolutely continuous (for example, the Cantor-Lebesgue function).
A: No, the function $f(x) = x \sin(1/x)$, $f(0)=0$ is not absolutely continuous
A: But function $x\sin\left(\frac{1}{x}\right)$ does not have bounded variation and it is continuous on $[0,1]$.
