In short, we do not need to adopt this as an axiom. But...
If there are sets at all, the axiom of subsets tells us that there is an empty set: If $x$ is a set, then $\{y\in x\mid y\ne y\}$ is a set, and is empty, since there are no elements $y$ of $x$ for which $y\ne y$. The axiom of extensionality then tells us that there is only one such empty set.
So, the issue is whether we can prove that there are any sets. The axiom of infinity tells us that there is a set (which is infinite, or inductive, or whatever formalization you use). But this seems like a terrible overkill to check that there are sets, to postulate that there are infinitely many.
Some people prefer to have an axiom that states that there are sets. Of course, some people then just prefer to have an axiom that states that there is an empty set, so we at once have that there are sets, and avoid having to apply comprehension to check that the empty set exists.
Other people adopt a formalization of first order logic in which we can prove that there are sets. More carefully, most formalizations of logic (certainly the one I prefer) prove as a theorem that the universe of discourse is nonempty. In the context of set theory, this means "there are sets". This is pure logic, before we get to the axioms of set theory. Under this approach, we do not need the axiom that states that there are sets, and the existence of the empty set can be established as explained above.
(The logic proof that there are sets is not particularly illuminating or philosophically significant. Usually, one of the axioms of first order logic is that $\forall x\,(x=x)$. If $\exists x\,(x=x)$ --the formal statement corresponding to "there are sets"-- is false, then $\forall x\,(x\ne x)$. Instantiating, we obtain $x\ne x$, and instantiating the axiom $\forall x\,(x=x)$ we obtain $x=x$, and one of these conclusions is the negation of the other, which is a contradiction. This is not particularly illuminating, because of course we choose our logical axioms and rules of instantiation so that this silly argument can go through, it is not a deep result, and probably we do not gain much insight from it.)
It turns out that yet some others prefer to allow the possibility that there are empty universes of discourse, so their formalization of first order logic is slightly different, and in this case, we have to adopt some axiom to conclude that there is at least one set.
At the end of the day, this is considered a minor matter, more an issue of personal taste than a mathematical question.
The reflection principle is a theorem schema in ZFC, meaning that for each formula $\phi(\vec x)$ we can prove in ZFC a version of the principle for $\phi$. In particular, it gives us that if $\phi$ holds (in the universe of sets) then there is some ordinal $\alpha$ such that $V_\alpha\models \phi$.
It follows from this that (assuming its consistency) $\mathsf{ZFC}$ is not finitely axiomatizable. Otherwise, $\mathsf{ZFC}$ would prove its own consistency, violating the second incompleteness theorem. The (standard) list of axioms you presented is actually an infinite list, with replacement being in fact an axiom schema (one axiom for each formula).
It is perhaps worth mentioning that no appeal to the incompleteness theorem is needed: If $\mathsf{ZFC}$ is consistent, and finitely axiomatizable, then it would prove (because of reflection) that there are $\alpha$ such that $V_\alpha\models\mathsf{ZFC}$. It would then follow that there is a least such $\alpha$. But inside $V_\alpha$ there must be some $\beta$ such that $$V_\alpha\models\mbox{``}V_\beta\models\mathsf{ZFC}\mbox{''}$$
(because $V_\alpha$ is a model of set theory, so it satisfies reflection), and easy absoluteess arguments give us that then $\beta<\alpha$ is indeed an ordinal, and $V_\beta$ is really a model of $\mathsf{ZFC}$, contradicting the minimality of $\alpha$.
Best Answer
With your edit, it is now clearer what you're asking -- but on the other hand you're now asking an extremely broad question that is basically the same as, "How do I prove anything in ZFC?"
Each and every axiom of ZFC -- with the exception of Extensionality and Regularity -- is there to claim that a set with certain properties exists. In order to prove that a set with whatever properties you're interested in exists, you combine those axioms into an argument and ends up with your desired conclusion. This can take ingenuity! There's no one-size-fits-all process that will allow you to go from a description of your desired set to a proof that exists without having to think and be clever along the way.
This is still a bit of a strange question, because usually saying that you have "constructed" something is exactly the same as saying that you have found a combination of the set existence axioms that lead to the conclusion that it exists. Speaking about "constructing" things is a more vivid way of saying it, but what you actually do is no different from putting together an existence proof.
To be concrete, if we assume that you already have $v_0$ and $v_1$:
Apply the Axiom of Pairing to $v_0$ and $v_0$ itself. This tells you that there exists a set whose members are exactly every $y$ that equals $v_0$ or equals $v_0$. By convention, this is the set we notate $\{v_0\}$ -- so the axiom says it exists!
Apply the Axiom of Pairing again, now to $v_0$ and $v_1$. It tells you that there exists a set whose members are exactly $v_0$ and $v_1$. In other words the notation $\{v_0,v_1\}$ describes something that exists.
Apply the Axiom of Pairing a third time, now to the sets $\{v_0\}$ and $\{v_0,v_1\}$ that we constructed just previously. Again, you conclude that a set whose members are precisely $\{v_0\}$ and $\{v_0,v_1\}$ exists. Which was what we set out to prove.
(To moderate the above claim a bit, for the nitpickers: Sometimes to "construct" something is taken to mean proving that it exists using a particularly restricted set of tools, i.e. no use of the Axiom of Choice and no use of the logical Law of the Excluded Middle. However, it is more common to use the adjective "constructive" for this restricted meaning, and still let the verb "construct" be used for every straightforward direct existence proof).