If $f'$ is bounded on finite subintervals, is it bounded on $[0,\infty)$? I am trying to show the following:

If $f$ is a bounded uniformly continuous function on $[0,\infty)$ and $f'$ is differentiable on $(0,\infty)$ such that $f'$ is bounded on each finite interval $(a,b)$, where $a,b\in [0,\infty)$, then $f'$ is bounded on $(0,\infty)$.

I had tried applying the Mean Value Theorem and proving it using contradiction but was not able to get the solution.
Any type of help will be appreciated.

Let $f$ be continuously differentiable such that:
For $n\in \Bbb N$
(i). $f(x)=0$ for $x\in [2n-1,2n-\frac {1}{2n^2}]\cup [2n+\frac {1}{2n^2},2n+1].$
(ii). $f$ is monotonic on $[2n-\frac {1}{2n^2},2n]$ and monotonic on $[2n,2n+\frac {1}{2n^2}].$
(iii). $f(2n)=1/n.$
Given $\epsilon>0,$ take $n\in \Bbb N$ large enough that  $\frac {1}{n+1}<\epsilon$ and take $\delta \in (0,\frac {1}{2n^2})$ small enough that $\forall x,y \in [0,2n+1]\,(\,|x-y|<\delta\implies |f(x)-f(y)|<\epsilon).$
Now if $|x-y|<\delta<\frac {1}{2n^2}$ and at least one of $x,y$ is greater than $2n+1$ then $\min (x,y)\ge 2n+\frac {1}{2n^2},\,$ so $|f(x)-f(y)|\le \max \{f(u): u\ge 2n+\frac {1}{2n^2}\}-\min \{f(v):v\ge 2n+\frac {1}{2n^2}\}=\frac {1}{n+1}<\epsilon.$
So $f$ is uniformly continuous.
For $n\in \Bbb N$ there exists $x_n\in (2n-1/2n^2,2n)$ such that $f'(x_n)= \frac {f(2n)-f(2n-1/2n^2)}{2n-(2n-1/2n^2)}=2n.$
So $f'$ is unbounded on $\Bbb R^+.$
