If $f_n^2\to f^2$ uniformly, with $f$ and $fn$ being non-negative, does $f_n\to f$ uniformly? I think it is true, and what I am trying to do is, since $f_n^2\to f^2$ uniformly, then for all $\epsilon > 0$, there exists $m\in \Bbb Z$ such that for all $x$, and for all integer $n\geq m$, we have $\left|f_n^2(x) - f^2(x)\right|<\epsilon$. I am not sure if I could split the problem into cases based on the value of $f(x)$ as following, if $f(x) = 0$, then |$f_n^2(x) - 0$| < $\epsilon^2$, which implies |$f_n(x) - 0$| < $\epsilon$, that is |$f_n(x) - f(x)$| < $\epsilon$, and if $f(x) \neq 0$, then let $M = f_n(x) + f(x)$, then we have |$f_n^2(x) - f^2(x)$| < $M\epsilon$, then |$f_n(x) - f(x)$| $=$ $\left|\frac{f_n^2(x) - f^2(x)}{f_n(x) + f(x)}\right| < \frac{M\epsilon}{M} = \epsilon$. I wonder if this approach is correct, is it OK to vary the $\epsilon$ base on the values of $x$ and $f(x)$? Is there any other better ways to prove this statement?

That is no correct. The number $\varepsilon$ remains fixed from the start.
Take $\varepsilon>0$. Since the square root function is uniformly continuous, there is some $\delta>0$ such that$$(\forall x,y\geqslant0):|x-y|<\delta\implies \left|\sqrt x-\sqrt y\right|<\varepsilon.$$So, take $N\in\Bbb N$ so large that$$(\forall n\in\Bbb N)(\forall x\geqslant0):n\geqslant N\implies\left|f_n^{\,2}(x)-f^2(x)\right|<\delta$$and then$$(\forall n\in\Bbb N)(\forall x\geqslant0):n\geqslant N\implies\left|f_n(x)-f(x)\right|=\left|\sqrt{f_n^{\,2}(x)}-\sqrt{f^2(x)}\right|<\varepsilon.$$
