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.
There seems to be no connection between $D$ and $A$ in $(*)$. So $(*)$ is correct. For one example take $D=[0, \frac 12 ], A=[0,1]$, $f_n(x)=x^{n}, f(x)=0$ for $ x<1$, $f(1)=1$. Then $f_n \to f$ uniformly on $D$ and $f$ is not continuous on $A$.
However, the example given by the lecturer is wrong since $f_n$ does not converge uniformly to $f$ on $D=\mathbb R$.
Answer for the revised question: If $f_n \to f$ uniformly on $A$ and each $f_n$ is continuous on $A$ then the restriction of $f$ to $A$ is continuous. But you cannot say that $f$ is continuous at points of $A$ as in the example by your lecturer.
Best Answer
For the second case, take $f_n(x)=g_n(x) = x+\frac{1}{n}$. With $f(x)=g(x) = x$, we see that $f_n \to f, g_n \to g$ uniformly.
However, $f_n(x) g_n(x) = f(x)g(x) + \frac{2}{n} x + \frac{1}{n^2}$, hence the convergence is not uniform on unbounded sets.