Question: Prove that the sequence of functions $f_n:(0,1)\to\mathbb{R}$ below is equicontinuous and simply converges to $f\equiv\dfrac{1}{2}$, but with not uniformly convergent subsequence in $(0,1)$. $$f_n= \begin{cases} \dfrac{1}{4}\sin\left(\dfrac{1}{x}\right)+\dfrac{1}{2}&,x\in\left(0,\dfrac{1}{2\pi n}\right]\\[1em] \quad\quad\quad\dfrac{1}{2}&,x\in\left[\dfrac{1}{2\pi n},1\right) \end{cases}.$$
My attempt: For the pointwise convergence, we have:
• $x=\dfrac{1}{2\pi n}$ $$f_n\left(\dfrac{1}{2\pi n}\right)=\dfrac{1}{4}\sin(2\pi n)+\dfrac{1}{2}=\dfrac{1}{2}.$$
• $x>\dfrac{1}{2\pi n}$ $$f_n(x)=\dfrac{1}{2}.$$
So, I think that we can conclude that $f_n\to f=\dfrac{1}{2}$ simply.
Now, my problem is to prove that the sequence of functions is equicontinuous with not uniformly convergent subsequence... what I was trying to do is to prove by the definition of equicontinuity:
A sequence of functions is equicontinuous in $x_0\in X$ when the set of functions $F=\{f_1,f_2,\dots,f_n,\dots\}$ is equicontinuos at $x_0$, in other words, for all $\varepsilon>0$, exists $\delta>0$ such that $$x\in X,|x-x_0|<\delta\Rightarrow|f_n(x)-f_n(x_0)|<\varepsilon,\forall~n\in\mathbb{N}.$$
and trying to find a $\delta$:
$$\left|\left[\dfrac{1}{4}\sin\left(\dfrac{1}{x}\right)+\dfrac{1}{2}\right]-\left[\dfrac{1}{4}\sin\left(\dfrac{1}{x_0}\right)+\dfrac{1}{2}\right]\right|=\left|\dfrac{1}{4}\left[\sin\left(\dfrac{1}{x}\right)-\sin\left(\dfrac{1}{x_0}\right)\right]\right|$$
but it didn't work. I don't know if I'm on the right way... and I also need to prove that the sequence of functions doesn't have a uniformly convergent subsequence in $(0,1)$.
Edit: I think I prove that $f_n$ is equicontinuous using the mean value theorem: $$\left|\dfrac{1}{4}\left[\sin\left(\dfrac{1}{x}\right)-\sin\left(\dfrac{1}{x_0}\right)\right]\right|\overset{\mbox{MVT}}{=}\dfrac{1}{4}\left|\dfrac{\cos\left(\frac{1}{c}\right)}{c}\cdot(x-x_0)\right|\leq\left|-\dfrac{1}{c}\right||x-x_0|<\dfrac{1}{c}\delta.$$
Now, choosing $\delta=c\cdot\varepsilon$ we have that $f_n$ is equicontinuous... but I still don't know how to prove that the sequence of functions doesn't have a uniformly convergent subsequence in $(0,1)$...