Fully-ordered set and transfinite mathematical induction 
Let $A$ be a fully-ordered class and let $P(x)$ be a statement which is either true or false for each element $x\in A$. Suppose that following condition holds:
Ind. If $P(y)$ is true for every $y<x$, then $P(x)$ is true.  
Then, $P(x)$ is true for every element $x \in A$.

Statement: If $A$ satisfies the above things, then $A$ is well-ordered.
I want to prove the statement, but I'm lost.
Suppose A is a fully-ordered set but not well-ordered, and A also satisfies the above things. Since A is not well-ordered, $\exists B\subseteq A$ such that $B$ does not have a least element. I want to show this leads to a contradiction.
 A: What you have lost when you give up the well order is that $A$ has a least element.  When there is one, your inductive statement says that $P$ is true of that element as there are no predecessors.  Without that, you could have $P$ false of all the elements.  The inductive statement is still true.  One example would be to take $A$ as the rationals and $P(x)$ to say $x$ is irrational.
A: In its current form your claim is wrong. What you probably want is:
Claim 1. Let $(A; <)$ be a strict, (set-like) linear order (where we allow $A$ to be a class) that is not well-ordered. Then there is some condition $P(x)$ such that if $P(y)$ is true for all $y < x$ then $P(x)$ is true and yet there is some $x \in A$ such that $P(x)$ is false.
In other words: The principle of mathematical induction fails for some property $P$ for $(A; <)$ and moreover $P$ can be chosen to be independent of $(A; <)$.
Proof. Let $P(x)$ be the statement "there is no infinite decreasing sequence $x > x_1 > x_2 > \ldots$.
Since $(A; <)$ is no well-founded we indeed have that $P(x)$ is false for some $x \in A$.
However, suppose that $P(y)$ holds for all $y < x$. I claim that $P(x)$ holds: Suppose not. Then there is some infinite decreasing sequence $x > x_1 > x_2 > \ldots$. But now $(x_2, x_3, \ldots)$ witnesses that $P(x_1)$ fails. Since $x_1 < x$ this is the desired contradiction. Q.E.D.
(Note that 'set-like' may not be necessary - depending on how you define well-orders.)
