Assume $(f_n)$ is a sequence of differentiable function. If $f_n\rightarrow f$ uniformly on $(-\infty, \infty)$, can we conclude that $f_n^2\rightarrow f^2$ uniformly on $(-\infty, \infty)$? If not, is there any counterexample?
I don't think this is true but having trouble of finding a counterexample. $|f_n^2-f^2|=|(f_n+f)(f_n-f)|$, so if it's uniformly convergent, the distance should be converging to $0$ depending on $n$. But $|f_n+f|$ may not be bounded, so I don't think the above statement could be true... Is that correct? Shouldn't the statement be true if it's on $[a,b]$ instead of $(-\infty, \infty)$?
You've figured out a weak point in a proof that it does converge uniformly. The next step is to look for counterexamples that exploit this, i.e. let $f$ and $f_n$ be unbounded functions. Might I suggest $$f_n(x)=x-\frac1n,\quad f(x)=x$$as probably the simplest one?
On a bounded, closed intervals, however, your proof works, because you can bound $|f+f_n|$ by, say, $$2\max_{x\in[a,b]}(|f|)+1$$ (which is a finite number) as long as $n$ is large enough.