This question below is inspired by this one. Based on the thought acquired from there, I was trying to solve the following question from Kesavan's Functional Analysis:
Here is my attempt:
For $n\in\mathbb N,$ define $f_n:\mathbb R\to\mathbb R$ by
$f_n(x)=\dfrac{1}{m}\sin x,$ if $(m-1)\pi\le x\le m\pi$ for some $m\in\{1,2,...,n\}$
$=0,$ otherwise
Choose $\epsilon>0.$ Then $\exists~p\in\mathbb N$ such that $\epsilon>\dfrac{1}{p+1}.$
Then for $m,n\ge p,$ $||f_m-f_n||=\sup_{x\in\mathbb R}|\dfrac{1}{\min\{m,n\}+1}|\le\dfrac{1}{p+1}<\epsilon$.
So $\{f_n\}$ is Cauchy.
If possible let $f_n\to f$ in $\mathcal C_{\mathcal C}(\mathbb R).$ Then $\text{supp}f\subset[a,b]$ for some $a,b\in R.$
Choose an odd natural number $k\in\mathbb N$ such that $k.\dfrac{\pi}{2}>b$ Then $f(k.\dfrac{\pi}{2})=0.$
But $f_n(k.\dfrac{\pi}{2})=\dfrac{1}{\frac{k-1}{2}+1}=\dfrac{2}{k+1}~\forall~n\ge k$ since $\frac{k-1}{2}\pi\le\frac{k}{2}\pi\le\frac{k+1}{2}\pi.$
So $||f_n-f||\ge|f_n(k.\dfrac{\pi}{2})-f(k.\dfrac{\pi}{2})|=\dfrac{2}{k+1}$ for all $n\ge k,$ a contradiction.
Hence $\mathcal C_{\mathcal C}(\mathbb R)$ is not complete.
Am I correct?
Your proof is fine. In fact, using your idea, I'd like to offer a streamlined version:
Take any function $0<f<1$, with support in $[0,1].$ A triangle will do, or if you want your functions to be differentiable, a bump function.
Now translate $f$ by defining $f_n(x)=f(x-n).$ Then each $f_n$ is supported in $[n,n+1]$ and the supports are disjoint.
Now just define $F_n(x)=\sum_{k=0}^n \frac{f_k(x)}{k+1}.$ Then, if $n>m,\ $ $|F_n(x)-F_m(x)|\le \frac{1}{m+1}$ so $(F_n)$ is Cauchy, but the limit function $F$ is clearly not compactly supported.