This problem from TIFR 2013, Problem 19 discusses the example of a non-uniformly continuous function.
Question: TIFR 2013 problem 19
True/False?
Every differentiable function $f:(0,1) \to [0,1]$ is uniformly continuous.
Hint:
$\sin(1/x)$ is not uniformly continuous. However, range is not $[0,1]$. But its a simple matter of scaling.
Discussion:
Let $f(x)=\sin(\frac{1}{x})$ for all $x\in(0,1)$. For simplicity, we first prove that this $f$ is not uniformly continuous. Then we will scale things down, which won't change the non-uniform continuity of $f$.
Note that $f$ is differentiable. To show that $f$ is not uniformly continuous, we first note that as $x$ approaches $0$, $\frac{1}{x}$ goes through an odd multiple of $\frac{\pi}{2}$ to an even multiple of $\frac{\pi}{2}$ real fast. So in a very small interval close to $0$, I can find two such points which gives value $1$ and $0$.
Let $x=\frac{2}{(2n+1)\pi}$ and $y=\frac{2}{2n\pi}$.
Then $|x-y|=\frac{2}{2n(2n+1)\pi}$.
Since the right hand side of above goes to zero as $n$ increases, given any $\delta > 0$ we can find $n$ large enough so that $|x-y|<\delta$. For these $x$ and $y$, $|f(x)-f(y)|=1$.
Of course, choosing $\epsilon$ as any positive number less than $1$ shows that $f$ is not uniformly continuous.
We have with us a function $f$ which is bounded, differentiable and not uniformly continuous.
To match with the questions requirements, notice $-1\le f(x)\le 1$.
So $0\le 1+f(x) \le 2$ And $0\le \frac{1+f(x)}{2} \le 1$.
Define $g(x)= \frac{1+f(x)}{2}$ for all $x\in(0,1)$.
Since sum of two uniformly continuous functions is uniformly continuous and a scalar multiple of uniformly continuous function is uniformly continuous, if $g(x)$ was uniformly continuous, then $f(x)=2g(x)-1$ would also be uniformly continuous.
This proves that g is in fact not uniformly continuous. It is still differentiable, and range is $[0,1]$, which shows that the given statement is actually false.