We can't really construct an infinite D-finite set by defining it, simply because it is consistent that there are no such sets (i.e. all the D-finite sets are finite).
We can construct models in which the axiom of choice fails and there are infinite D-finite sets, but those constructions are difficult and involve forcing in many cases.
First note that a finite set is D-finite. So to solve the second question means to solve the first one as well.
To show that if $A$ and $B$ are D-finite then $A\cup B$ is also D-finite, assume by contradiction that $A\cup B$ is not D-finite then there is some $X\subseteq A\cup B$ such that $|X|=\aleph_0$. Consider $X\cap A$ and $X\cap B$, and show that at least one of those has size $\aleph_0$, in contradiction to the assumption that both sets are D-finite.