Is $\{x_n\}$ a Cauchy sequence? Let $(X,d)$ be a complete metric space and $f$ a mapping of $X$ into itself. Let $\{f^n(x)\}=\{x_n\}$ be the sequence of iterated transforms.
Suppose $f$ satisfies that for each $\varepsilon >0$,there exists $\delta>0$ such that for all $x,y\in X$,
$$\varepsilon\leq d(x,y)<\varepsilon+\delta\implies d(f(x),f(y))<\varepsilon.$$
Is $\{x_n\}$ a Cauchy sequence ?
 A: $\let\eps\varepsilon$Yes, that's true, and the argument is somewhat similar to that for your previous question. Surely, completeness is not needed, since you wish to obtain a Cauchy sequence, not a convergent one.
Denote $d_i=d(x_i,x_{i+1})$. Applying the condition for $\eps=d_i$, we get that $d_i$ strictly decrease, so there exists $D=\lim_{i\to\infty} d_i$. If $D>0$, we get a contradiction by setting $\eps=D$; so $D=0$.
Now let us show that for every $\eps>0$ there exists $N$ such that for all $k\geq n>N$ we have $d(x_k,x_n)<2\eps$; this is exactly what we need. Choose $\delta$ for our $\eps$ from the condition; we may assume that $\delta<\eps$. There exists $N$ such that $d_n<\delta$ for all $n>N$. We claim that this $N$ fits.
Indeed, let us show that for all $k\geq n>N$ we have $d(x_k,x_n)<\eps+\delta$; this is clearly sufficient. Induction on $k$. If $k=n$ then there is nothing to prove. Assume now that $k>n$. [ADDED] By the induction hypothesis, we have $d(x_{k-1},x_n)<\eps+\delta$. Now consider the following two cases.[/ADDED]
If $d(x_{k-1},x_n)<\eps$ then 
$$
  d(x_k,x_n)\leq d_{k-1}+d(x_{k-1},x_n)<\delta+\eps.
$$ 
Otherwise, we have $\eps \leq d(x_{k-1},x_n)<\eps+\delta$, so by the condition $d(x_k,x_{n+1})<\eps$, and 
$$
  d(x_k,x_n)\leq d(x_k,x_{n+1})+d_n\leq \eps+\delta.
$$
The claim is proved.
