Comparison
Pointwise convergence means at every point the sequence of functions has its own speed of convergence (that can be very fast at some points and very very very very slow at others).
Imagine how slow that sequence tends to zero at more and more outer points:
$$\frac{1}{n}x^2\to 0$$
Uniform convergence means there is an overall speed of convergence.
In the above example no matter which speed you consider there will be always a point far outside at which your sequence has slower speed of convergence, that is it doesn't converge uniformly.
Another Approach
One can check uniform convergence by considering the "infimum of speeds over all points". If it doesn't vanish then it is uniformly convergent. And that gives another characterization as the ones with nonvanishing overall speed of convergence.
Consider the set of all bounded functions from $[a,b]$ to $\mathbb{R}$ which we denote by $B([a,b],\mathbb{R})$. Defining a norm for this set as the usual Sup norm
$$\left\|f\right\|=\text{Sup}\{|f(x)|:x\in[a,b]\}$$
and considering its induced metric
$$d(f,g)=\text{Sup}\{|f(x)-g(x)|:x\in[a,b]\}$$
one can show that this will be a metric space (check this as an exercise). Now, we claim that convergence in this metric space is equivalent to uniform convergence on $[a,b]$. To check this out, let us write the definitions as follows
\begin{align*}
&\text{Convergent in $B([a,b],\mathbb{R})$} \iff \\
&\forall\epsilon_1\gt0,\,\,\exists N_1\gt0,\,\,n\ge N_1 \implies \text{Sup}\{|f_n(x)-f(x)|,x\in[a,b]\}<\epsilon_1 \\
\\
&\text{Uniformly Convergent on $[a,b]$} \iff \\
&\forall\epsilon_2\gt0,\,\,\exists N_2\gt0,\,\, \forall x\in [a,b],\,\, n\ge N_2 \implies |f_n(x)-f(x)|<\epsilon_2.
\end{align*}
First assume convergence in $B([a,b],\mathbb{R})$. Consequently, $\epsilon_2$ is given and we should find $N_2$ such that the desired result holds. Choose $\epsilon_1:=\epsilon_2$ and take $N_2:=N_1$. Then according to
$$|f_n(x)-f(x)|\le\text{Sup}\{|f_n(x)-f(x)|,x\in[a,b]\}<\epsilon_2,\qquad \forall x\in [a,b]$$
uniform convergence on $[a,b]$ follows immediately. Next, assume uniform convergence on $[a,b]$. So $\epsilon_1$ is given and we should find $N_1$ such that the result holds. Choose $\epsilon_2:=\frac{\epsilon_1}{2}$ and take $N_1:=N_2$. For fixed $n$, $\frac{\epsilon_1}{2}$ is an upper bound for $\{|f_n(x)-f(x)|\,\big|x\in[a,b]\}$ and due to the completeness of $\mathbb{R}$ this set must have a supremum, which is smaller than all of its upper bounds. This leads us to
$$\text{Sup}\{|f_n(x)-f(x)|,x\in[a,b]\}\le\frac{\epsilon_1}{2}\lt\epsilon_1$$
which completes the proof.
Best Answer
we have that $f_{n} \to 0$ uniformly on $E$. Then $\sup|f_{n}(x)| \to 0$. Thus for $\varepsilon > 0$ there exists some $N>0$ such that if $n\geq N \implies |f_{n}(x)| \leq \sup|f_{n}(x)|< \varepsilon $ for all $x \in E$. Now, take $\varepsilon > 0$, and $\tilde{\varepsilon} = \frac{\varepsilon}{M}$. Then there exist some $N > 0$ such that if $n \geq N \implies |f_{n}(x)| < \tilde{\varepsilon}$ for all $x \in E$. Then $|f_{n}(x)g_{n}(x)| = |f_{n}(x)||g_{n}(x)| \leq |f_{n}(x)|M < \tilde{\varepsilon}M = \varepsilon$ for all $n\geq N$, for all $x \in E$.
Then $\sup|f_{n}(x)g_{n}(x)| \to 0$
So $f_{n}g_{n} \to 0$ uniformly.