Consider the function $f\colon (0,1]\to\mathbb{R}$ given by $$ f(x):=\begin{cases}n(n+1)x-n, & \textrm{if }x\in\left[\frac{1}{n+1},\frac{1}{n}\right], n\in\mathbb{N}\textrm{ odd}\\-n(n+1)x+n+1, & \textrm{if }x\in\left[\frac{1}{n+1},\frac{1}{n}\right], n\in\mathbb{N}\textrm{ even}\end{cases} $$ and show that it is bounded and continuous, but not uniformly continuous.
The first thing I am dealing with is to understand the definition of $f$. I try to simplify it.
When I am not mistaken, then $$ f(x)=2x-1,\qquad\frac{1}{2}< x\leq 1 $$
and I am trying to simplifiy $f$ also on $\left( 0 , \frac{1}{2}\right]$. I already found out that $$ f\left(\frac{1}{2}\right)=0. $$
Maybe it is not necessary to simplify $f$ but in the form it is given, it is hard for me to analyse it.
Hint: fix an odd n. Then on the nth interval, the function is a linear function from 0 to 1 (to see this, substitute the endpoints into the equation of f). Now, n+1 is even, and we similarly see that on the n+1st interval the function is a line segment from 1 to 0. Finally, notice that as n increases, the intervals become narrower and closer to 0.
From this, can you tell what the graph of f looks like?