Continuity on infinity We have $f: [a.b] \rightarrow \mathbb{R}$ continuous, and 
$$c_n = \sup\{c \in [a,b] : |f(x) - f(c_{n-1})| < \epsilon \text{ for any } x \in [c_{n-1},c]\}$$
with $c_1 = a$. In a previous exercise we saw that $c_n$ goes to $b$, so we have those intervals $[c_{n-1},c]$ covering the whole of $[a,b]$.
What I have been tasked to do is to prove, using this, that a continuous function on a closed and bounded interval is uniformly continuous. I have some clue about how to go about this, yet I cannot figure out how to prove this is the case if for some $x$ and $\delta$, $x-\delta$ is on one of the intervals and $x+\delta$ is on another.
 A: You've shown that for any given $\epsilon>0$, you have a corresponding partition $\{c_n\}_{n=1}^{N}$ of the interval $[a,b]$ such that $c_1=a,$ $ c_N=b$, and for each $1 \leq n < N$, $\forall x \in [c_n,c_{n+1})$ you have $|f(x)-f(c_n)| < \epsilon$. Note this implies also that $|f(c_{n+1})-f(c_n)| \leq \epsilon$.
Fix an $\epsilon>0$, and consider the above sequence $\{c_n\}$ corresponding to $\frac{\epsilon}{3}$. Choose 
$$\delta= \min_{1 \leq n < N}\{c_{n+1}-c_{n}\}$$
Then for any $x,y \in [a,b]$ such that $|y-x| < \delta$, suppose WLOG that $x \leq y$ and consider that $\exists$ $1 \leq n < N$ such that $x \in [c_n,c_{n+1}]$, and we either have $y \leq c_{n+1}$ or $y > c_{n+1}$.
In the former case, $y \in [c_n,c_{n+1}]$, so 
$$|f(y)-f(x)| \leq |f(y)-f(c_n)|+|f(x)-f(c_n)| \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} < \epsilon$$
By construction of the partition.
In the latter case, $y > c_{n+1} \geq x$ (note $n \leq N-2$ for this case to occur), so
$$y-c_{n+1} \leq y-x < \delta \leq c_{n+2}-c_{n+1}$$
and hence $y < c_{n+2}$, i.e. $y \in [c_{n+1},c_{n+2})$, and we have
$$|f(y)-f(x)| \leq |f(y)-f(c_{n+1})| + |f(c_{n+1})-f(c_n)| + |f(x)-f(c_n)| < \frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3} = \epsilon$$
again by construction of the partition, showing that $f$ is uniformly continuous on $[a,b]$.
