Construct sequences $\{f_n\}$, $\{g_n\}$ which converge uniformly in some set $E$, but such that $\{f_ng_n\}$ does not converge uniformly on $E$ (of course, $\{f_ng_n\}$ must converge on $E$).
My example: Let's $E=(0,1)$ and $f_n(x)=\dfrac{1}{x}, g_n(x)=\dfrac{1}{n+x^{-1}}.$
Then $\{f_n\}$ converges uniformly on $E$ since $M_n=\sup\limits_{E}|f_n(x)-f(x)|=0$. Also $\{g_n\}$ converges uniformly on $E$ since $|g_n(x)|=\dfrac{1}{n+x^{-1}}\le \dfrac{1}{n}\to 0$ as $n\to \infty$.
But $f_n(x)g_n(x)=\dfrac{1}{nx+1}$ does not converges unifromly on $E$ since $M_n=\sup\limits_{E}|f_n(x)g_n(x)|\ge f_n(\frac{1}{n})g_n(\frac{1}{n})=\frac{1}{2}.$
What you think about above example? I would be grateful to anyone's opinion.
What you proved is that the sequence $(f_ng_n)$ cannot converge uniformly to $0$. Maybe we then need an argument which say that since $f_ng_n\to 0$ pointwise, then the unique potential uniform limit is the null function.
Otherwise, your counter-example is fine.