Since $A\subset \mathbb R^n$, then in order to show that $A$ is compact, it suffices to show that $A$ is bounded and closed.
$A$ is bounded. Clearly the function $f:A\to\mathbb R$, with $f(x)=\|x\|$ is continuous, and as it has to be bounded, then there exists an $M>0$, such that $f(x)=\|x\|\le M$, for all $x\in A$. But this means that $A$ is bounded.
$A$ is closed. If not, then there exists a sequence $\{x_n\}_{n\in\mathbb N}\subset A$, with $x_n\to x_0\not\in A$. Then the function $\,f:A\to\mathbb R$, with
$$
f(x)=\frac{1}{\|x-x_0\|},
$$
in continuous and unbounded. A contradiction, and hence $A$ is closed.
Altogether, indeed $A$ is compact.
Note. Another way to obtain contradiction, is by observing that the function $g(x)=\|x-x_0\|$ is also continuous, but it does not attain a minimum. The infimum of $f$ is equal to zero, while $f$ takes only positive values.
Yes, your set is bounded. But its not closed, since its complement is $(-\infty,0)\cup(1,2]\cup(3,+\infty)$, which is not open: $2$ belongs to it, but no open interval centered at $2$ is contained in it.
Best Answer
Note that $\overline D\subset\overline A=A$. Besides, since $D$ is bounded, $\overline D$ is bounded too. So, $\overline D$ is a compact subset of $A$. Therefore, $f\left(\overline D\right)$ is compact too, and so it is bounded. In particular, $f(D)$ is bounded, and so, $\overline{f(D)}$ is bounded too. Sinse it is a closure, it is also a closed set.