Okay, I can't resist: here is a quick answer.
I am construing the question in the following way: "Is there some criterion for a subset of $[0,\infty)$ to be all of $[0,\infty)$ which is (a) analogous to the principle of mathematical induction on $\mathbb{N}$ and (b) useful for something?"
The answer is yes, at least to (a).
Let me work a little more generally: let $(X,\leq)$ be a totally ordered set which has
$\bullet $a least element, called $0$, and no greatest element.
$\bullet$ The greatest lower bound property: any nonempty subset $Y$ of $X$ has a greatest lower bound.
Principle of Induction on $(X,\leq)$: Let $S \subset X$ satisfy the following properties:
(i) $0 \in S$.
(ii) For all $x$ such that $x \in S$, there exists $y > x$ such that $[x,y] \subset S$.
(iii) If for any $y \in X$, the interval $[0,y) \subset S$, then also $y \in S$.
Then $S = X$.
Indeed, if $S \neq X$, then the complement $S' = X \setminus S$ is nonempty, so has a greatest lower bound, say $y$. By (i), we cannot have $y = 0$, since $y \in S$. By (ii), we cannot have $y \in S$, and by (iii) we cannot have $y \in S'$. Done!
Note that in case $(X,\leq)$ is a well-ordered set, this is equivalent to the usual statement of transfinite induction.
It also applies to an interval in $\mathbb{R}$ of the form $[a,\infty)$. It is not hard to adapt it to versions applicable to any interval in $\mathbb{R}$.
Note that I believe that some sort of converse should be true: i.e., an ordered set with a principle of induction should have the GLB / LUB property. [Added: yes, this is true. A totally ordered set with minimum element $0$ satisfies the principle of ordered induction as stated above iff every nonempty subset has an infimum.]
Added: as for usefulness, one can use "real induction" to prove the three basic Interval Theorems of honors calculus / basic real analysis. These three theorems assert three fundamental properties of any continuous function $f: [a,b] \rightarrow \mathbb{R}$.
First Interval Theorem: $f$ is bounded.
Inductive Proof: Let $S = \{x \in [a,b] \ | \ f|_{[a,x]} \text{ is bounded} \}$. It suffices to show that $S = [a,b]$, and we prove this by induction.
(i) Of course $f$ is bounded on $[a,a]$.
(ii) Suppose that $f$ is bounded on $[a,x]$. Then, since $f$ is continuous at $x$, $f$ is bounded near $x$, i.e., there exists some $\epsilon > 0$ such that $f$ is bounded on
$(x-\epsilon,x+\epsilon)$, so overall $f$ is bounded on $[0,x+\epsilon)$.
(iii) If $f$ is bounded on $[0,y)$, of course it is bounded on $[0,y]$. Done!
Corollary: $f$ assumes its maximum and minimum values.
Proof: Let $M$ be the least upper bound of $f$ on $[a,b]$. If $M$ is not a value of $f$,
then $f(x)-M$ is never zero but takes values arbitrarily close to $0$, so $g(x) = \frac{1}{f(x)-M}$ is continuous and unbounded on $[a,b]$, contradiction.
(Unfortunately in the proof I said "least upper bound", and I suppose the point of proofs by induction is to remove explicit appeals to LUBs. Perhaps someone can help me out here.)
Second Interval Theorem (Intemediate Value Theorem): Suppose that $f(a) < 0$ and $f(b) > 0$. Then there exists $c \in (a,b)$ such that $f(c) = 0$.
Proof: Define $S = \{x \in [a,b] \ | \ f(x) \leq 0\}$. In this case we are given that $S \neq [a,b]$, so at least one of the hypotheses of real induction must fail. But which?
(i) Certainly $a \in S$.
(iii) If $f(x) \leq 0$ on $[a,y)$ and $f$ is continuous at $y$, then we must have $f(y) \leq 0$ as well: otherwise, there is a small interval about $y$ on which $f$ is positive.
So it must be that (ii) fails: there exists some $x \in (a,b)$ such that $f \leq 0$ on
$[a,x]$ but there is no $y > x$ such that $f \leq 0$ on $[a,y]$. As above, since $f$ is continuous at $x$, we must have $f(x) = 0$!
Third Interval Theorem: $f$ is uniformly continuous.
(Proof left to the interested reader.)
Moreover, one can give even easier inductive proofs of the following basic theorems of analysis: that the interval $[a,b]$ is connected, that the interval $[a,b]$ is compact (Heine-Borel), that every infinite subset of $[a,b]$ has an accumulation point (Bolzano-Weierstrass).
Acknowledgement: My route to thinking about real induction was the paper by I. Kalantari "Induction over the Continuum".
His setup is slightly different from mine -- instead of (ii) and (iii), he has the single axiom that $[0,x) \subset S$ implies there exists $y > x$ such that $[0,y) \subset S$ -- and I am sorry to say that I was initially confused by it and didn't think it was correct. But I corresponded with Prof. Kalantari this morning and he kindly set me straight on the matter.
For that matter, several other papers exist in the literature doing real induction (mostly in the same setup as Kalantari's, but also with some other variants such as assuming (ii) and that the subset $S$ be closed). The result goes back at least to a 1922 paper of Khinchin, who in fact later used real induction as a basis for an axiomatic treatment of real analysis. It is remarkable to me that this appealing concept is not more widely known -- there seems to be a serious PR problem here.
Added Later: I wrote an expository article on real induction soon after giving this answer: it's very similar to what I wrote above, but longer and more polished. It is available here if you want to see it.
About question n°1 :
Who coined the expression "mathematical induction"?
the qualificative "mathematical" was introduced in order to separate this method of proof from the inductive reasoning used in empirical sciences (the "all ravens are black" example); it is common also to call it complete induction, compared to the "incomplete" one used in empirical science.
The reason is straightforward : the mathematical method of proof establish a "generality" ("all odd numbers are not divisible by two") that holds without exception, while the "inductive generalization" established by observation of empirical facts can be subsequently falsified finding a new counter-example.
Note : induction (the non-mathematical one) was already discussed by Aristotle :
Deductions are one of two species of argument recognized by Aristotle. The other species is induction (epagôgê). He has far less to say about this than deduction, doing little more than characterize it as “argument from the particular to the universal”. However, induction (or something very much like it) plays a crucial role in the theory of scientific knowledge in the Posterior Analytics: it is induction, or at any rate a cognitive process that moves from particulars to their generalizations, that is the basis of knowledge of the indemonstrable first principles of sciences.
For the history of the name "mathematical induction", see
The process of reasoning called "mathematical induction" has had several independent origins. It has been traced back to the Swiss Jakob (James) Bernoulli [Opera, Tomus I, Genevae, MDCCXLIV, p. 282, reprinted from Acta eruditorum, Lips., 1686, p. 360. See also Jakob Bernoulli's Ars conjectandi, 1713, p. 95], the Frenchmen B.Pascal [OEuvres completes de Blaise Pascal, Vol. 3, Paris, 1866, p. 248] and P.Fermat [Charles S Peirce in the Century Dictionary, Art."Induction," and in the Monist, Vol. 2, 1892, pp. 539, 545; Peirce called mathematical induction the "Fermatian inference"], and the Italian F.Maurolycus [G.Vacca, Bulletin Am. Math. Soc., Vol. 16, 1909, pp. 70-73].
The process of Fermat differs somewhat from the ordinary mathematical induction; in it there is a descending order of progression, leaping irregularly over perhaps several integers from $n$ to $n - n_1, n - n_1 - n_2$, etc. Such a process was used still earlier by J.Campanus in his proof of the irrationality of the golden section, which he published in his edition of Euclid (1260).
John Wallis, in his Arithmetica infinitorum (Oxford, 1656), page 15, [uses] phrases like "fiat investigatio per modum inductionis" [...].He speaks, p. 33, of "rationes inductione repertas" and freely relies upon incomplete "induction" in the manner followed in natural science.
Thus, his method has been criticized by Fermat as being "conjectural", i.e.based on a perceived regularity or repeated schema in a group of formuale.
Wallis states (page 306) that Fermat "blames my demonstration by Induction, and pretends to amend it. . . . I look upon Induction as a very good method of Investigation; as that which doth very often lead us to the easy discovery of a General Rule."
For about 140 years after Jakob Bernoulli, the term "induction" was used by mathematicians in a double sense: (1) "Induction" used in mathematics in the manner in which Wallis used it; (2) "Induction" used to designate the argument from $n$ to $n + 1$. Neither usage was widespread. The former use of "induction" is encountered, for instance, in the Italian translation (1800) of Bossut and Lalande's dictionary,' article "Induction (term in mathematics)." The binomial formula is taken as an example; its treatment merely by verification, for the exponents $m = 1, m = 2, m = 3$, etc., is said to be by "Induction." We read that "it is not desirable to use this method, except for want of a better method." H.Wronski (1836) in a similar manner classed "methodes inductionnelles" among the presumptive methods ("methodes presomptives") which lack absolute rigor.
The second use of the word "induction" (to indicate proofs from $n$ to $n + 1$) was less frequent than the first. More often the process of mathematical induction was used without the assignment of a name. In Germany A.G.Kastner (1771) uses this new "genus inductionis" in proving Newton's formulas on the sums of weakness of Wallis's Induction, then explains Jakob Bernoulli's proof from $n$ to $n + 1$, but gives it no name. In England, Thomas Simpson [Treatise of Algebra, London, 1755, p. 205.] uses the $n$ to $n + 1$ proof without designating it by a name, as does much later also George Boole [Calculus of Finite Differences, ed. J.F.Moulton, London, 1880, p. 12.]
A special name was first given by English writers in the early part of the nineteenth century. George Peacock, in his Treatise on Algebra, Cambridge, 1830, under permutations and combinations, speaks (page 201) of a "law of formation extended by induction to any number," using "induction," as yet in the sense of "divination." Later he explains the argument from $n$ to $n + 1$ and calls it "demonstrative induction" (page 203).
The next publication is one of vital importance in the fixing of names; it is Augustus De Morgan's article "Induction (Mathematics)" in the Penny Cyclopedia, London, 1838. He suggests a new name, namely "successive induction," but at the end of the article he uses incidentally the term "mathematical induction." This is the earliest use of this name that we have seen.
Best Answer
It is possible indeed, the principle of induction is much more general than you may think. There are many induction theorems you can prove for the most intuitive of them if you understand the principal idea which is well-foundedness, or well-orderedness. (but maybe it is too soon for you to benefit from the effort you would have to make to understand these notions)
The usual induction theorem for $\mathbb{Z}$ can be proven using regular induction. It can be stated as:
If a property $P(x)$ satisfies $P(n)$ for some integer $n$ and $\forall k \in \mathbb{Z}(P(k) \longrightarrow P(k-1) $ and $P(k+1))$, then it is true for every integer.
If you don't know how to prove PMI using the fact that every non-empty subset of $\mathbb{N}$ has a least element, I suggest you look for a proof. I remember I was very happy to discover this at the time I thought PMI was just some kind of common belief among mathematicians; and it is still a very interesting theorem for me.