Characterization properties of number sets $\mathbb{N},\mathbb{ Z},\mathbb{Q},\mathbb{R},\mathbb{C}$
$\Bbb N$ is the unique linearly ordered set that is infinite, but every initial segment is finite; it is also the smallest set which contains $0,1$ and closed under addition that satisfies the axiom $x+y=0\iff x=y=0$.
$\Bbb Z$ is the smallest linearly ordered set on which both successor and predecessor operations are defined everywhere, and every element is a successor.
$\Bbb Q$ is the smallest field which satisfies that $1+1+\ldots+1\neq 0$. It is also the smallest field which can be ordered.
$\Bbb R$ is the unique ordered field which is Dedekind complete as an ordered set.
$\Bbb C$ is the unique field which is algebraically closed, contains $\Bbb Q$ and is equipotent with $\Bbb R$. It is also the unique algebraic closure of $\Bbb R$.
Most of these things are distinguished as being initial objects in some suitable category. I don't have much practice with all of the statements, so I apologize for missteps. The notion of being "initial" is mostly what "being the smallest" means, as in Asaf's answer.
When they are distinguished as initial objects of categories of ring-like objects with orders
I think $\Bbb N$ must be the initial object of the category of totally ordered semirings (with identity). (Morphisms would have to preserve both $0$ and $1$.)
$\Bbb Z$ is initial in the category of totally ordered rings. (We have given up well-ordering from $\Bbb N$.)
$\Bbb Q$ is initial in the category of totally ordered fields. (We have given up "discreteness" from $\Bbb Z$ in the following sense: given any point of the ordered set, there isn't a least point strictly above it or a greatest point strictly beneath it.)
$\Bbb R$ is initial in the category of totally ordered fields with the least upper bound property, but it turns out to be a let-down since it is the only such ring. (We have given up "holes" that existed in $\Bbb Q$. Every subset now has a least upper bound. )
When we get to $\Bbb C$ we have a problem because $\Bbb C$ isn't orderable. I'm not aware of any categorically reasonable way to continue describing $\Bbb C$ with the program of categories with ordered objects. $\Bbb R$ seems to have reached an apex of total-orderedness and continuity, and it looks like $\Bbb C$ has gone out of bounds.
What if we just pay attention to order alone?
$\Bbb N$ is initial in the category of sets with a distinguished point and a successor function. (This is an exercise in MacLane's CFTWM.)
I'm not certain, but I think $\Bbb Z$ is initial in the category of sets with a distinguished point and a successor function and a predecessor function. (I'll see if a variant of MacLane's exercise works out for this.) After this point we abandon successors and predecessors and switch to plain order.
We might hope that $\Bbb Q$ is initial in the category of totally ordered sets with a distinguished point; however, I'm not entirely convinced this is true. It feels like even if we had an order preserving and distinguished-point preserving map from $\Bbb Q$ to another set, maybe you can "scale" the map to get a different one. (When we are working with rings, the extra algebraic structure eliminates this problem.)
$\Bbb R$ is in a similar state as $\Bbb Q$, since it's not clear that it's initial in the category of complete totally ordered sets.
$\Bbb C$ even lacks a natural order, and we would still have problems similar to $\Bbb Q$ and $\Bbb R$.
What if we just pay attention to the ring-like structure?
$\Bbb N$ is initial in the category of semirings (with identity).
$\Bbb Z$ is initial in the category of rings (with identity).
$\Bbb Q$ is initial in the category of characteristic $0$ fields.
it doesn't seem likely that $\Bbb R$ has a purely algebraic characterization, considering its construction inherently references topological properties.
Given $\Bbb R$, then $\Bbb C$ does have the algebraic distinction of being the only algebraic field extension of $\Bbb R$ other than $\Bbb R$ itself. There is indeed a category of field extensions of $\Bbb R$, but this time $\Bbb C$ is not initial. Actually it seems to be terminal in this category. Apparently then it is initial in the opposite category :)
Finally, given any commutative ring $K$ (like a field), $K$ is bound to be initial in the category of associative $K$ algebras with identity.