Can you, please, check if the following proof is correct? Thanks for your time and effort.
Suppose that $f_n\to f$ and $g_n\to g$, as $n\to \infty,$ uniformly on $E\subseteq \Bbb{R}.$ Then, $f_n g_n\to fg,$ as $n\to \infty,$ pointwise on $E$.
Let $\epsilon>0$ be given and $x\in E$ 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 $\{f_n(x) \}_{n\in\Bbb{N}}$ and $\{g_n(x) \}_{n\in\Bbb{N}}$ are real sequences that converge to $f(x)$ and $g(x)$, respectively. So, they are bounded, i.e., for fixed $x\in E,$ 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 \;n\in\Bbb{N}.\end{align}
Therefore,
\begin{align}\lim\limits_{n\in\Bbb{N}}\left|g_n(x)\right| =\left|\lim\limits_{n\in\Bbb{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 $n\to \infty,$ for fixed $x\in E$, then there exists $N_1(\epsilon),N_2(\epsilon)$ such that
\begin{align} \left|f_n(x)-f(x) \right|<\dfrac{\epsilon}{2K} ,\;\;\forall\;n\geq N_1(\epsilon)\end{align}
and \begin{align} \left|g_n(x)-g(x) \right|<\dfrac{\epsilon}{2M } ,\;\;\forall\;n\geq 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\;n\geq N(\epsilon)\end{align}
Best Answer
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:
$$\lim_{n\to\infty} a_n \times \lim_{n\to\infty} b_n = \lim_{n\to\infty} a_n\times b_n$$
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.