Ytukyg

Can you, please, check if the following proof is correct? Thanks for your time and effort.

Suppose that $f_nto f$ and $g_nto g$, as $nto infty,$ uniformly on $Esubseteq Bbb{R}.$ Then, $f_n g_nto fg,$ as $nto infty,$ pointwise on $E$.

Let $epsilon>0$ be given and $xin E$ be fixed. Then,

begin{align} left|(f_n g_n)(x)-(f g)(x)right| &= left|f_n(x) g_n(x)-f(x) g(x)right| \&= left|f_n(x) g_n(x)-f_n(x) g(x)+f_n(x) g(x)-f(x) g(x)right| \&leq left|f_n(x) g_n(x)-f_n(x) g(x)right| + left|f_n(x) g(x)-f(x) g(x)right| \&= left|f_n(x)right| left|g_n(x)- g(x)right| + left|g(x)right| left|f_n(x)-f(x) right| end{align}
The sequences ${f_n(x) }_{ninBbb{N}}$ and ${g_n(x) }_{ninBbb{N}}$ are real sequences that converge to $f(x)$ and $g(x)$, respectively. So, they are bounded, i.e., for fixed $xin E,$ there exists $M>0$ and $K>0,$ such that begin{align}left|f_n(x)right| leq M;text{and};left|g_n(x)right| leq K,;forall ;ninBbb{N}.end{align}
Therefore,
begin{align}limlimits_{ninBbb{N}}left|g_n(x)right| =left|limlimits_{ninBbb{N}}g_n(x)right|=left|g(x)right|leq K.end{align}
So, we have begin{align} left|(f_n g_n)(x)-(f g)(x)right| &leq left|f_n(x)right| left|g_n(x)- g(x)right| + left|g(x)right| left|f_n(x)-f(x) right|\&leq M left|g_n(x)- g(x)right| + K left|f_n(x)-f(x) right|end{align}
Since $f_n(x)to f(x)$ and $g_n(x)to g(x)$, as $nto infty,$ for fixed $xin E$, then there exists $N_1(epsilon),N_2(epsilon)$ such that
begin{align} left|f_n(x)-f(x) right|<dfrac{epsilon}{2K} ,;;forall;ngeq N_1(epsilon)end{align}
and begin{align} left|g_n(x)-g(x) right|<dfrac{epsilon}{2M } ,;;forall;ngeq N_2(epsilon)end{align}
Let $N(epsilon)=max{N_1(epsilon),N_2(epsilon)}.$ Then,
begin{align} left|(f_n g_n)(x)-(f g)(x)right| &leq M left|g_n(x)- g(x)right| + K left|f_n(x)-f(x) right|\&<dfrac{epsilon}{2 } +dfrac{epsilon}{2 }=epsilon,;;forall;ngeq N(epsilon)end{align}

As I mentioned in the comments, you have not used the uniform convergence of $f_n$ or $g_n$ in your argument. And in fact, it is not necessary for your statement to hold. The important difference between pointwise and uniform convergence is that the later is global in the sense that the $N$ you find depends only on $epsilon$ and not on $x$. Geometrically, suppose that $f_n to f$ uniformly. Then, consider the graph of $f$ engulfed by $f+epsilon$ and $f-epsilon$. Picture this like a cylinder of radius $epsilon$ wrapped around $f$ at each point of its domain entirely.

When the convergence is uniform, after a certain $N$, all $f_n$'s will be contained in this tubular (cylindrical) neighborhood globally for $n geq N$. When the convergence is not uniform, no matter what $N$ is, some $f_n$'s will not be completely contained in this tubular (cylindrical) neighborhood. This is the geometric idea. In other words, the cylinder wrapped around $f$ might get wider or narrower around some points to contain $f_n$'s. It won't look uniform to us anymore.

In your case, your theorem is implied by the following simple theorem about real sequences:

$$lim_{ntoinfty} a_n times lim_{ntoinfty} b_n = lim_{ntoinfty} a_ntimes b_n$$

where $a_n$ and $b_n$ are real sequences. Note that for any $x$, $f_n(x)$ and $g_n(x)$ are real sequences. Now try to continue your reasoning from here.