If we assume that the axiom of choice is right, the well-ordering theorem can be verified. So, the set of real numbers also can be constructed by well-ordering property. By the well-ordering property, the entire set of real number has least element as x1(because the entire set of real number is the non-empty subset of real number) Then, well-ordering property also guarantees that the successor of non-greatest element of well-ordered set. So, we can choose the successor of x1 as x2. And the successor of x2 as x3 and so on. Then, we can label the entire element of real numbers as x1 x2 x3 ... This is obviously contradiction with the fact that the cardinality of real numbers is larger than that of natural numbers.
So I am very confused. What is the error that I missed? Help me, please.
Correct.
Incorrect. Just because you can enumerate some real numbers, it doesn't mean that such procedure will cover all of them.
For a simpler example, consider the natural numbers $\mathbb{N}$ together with additional element, which we will denote by $\infty$. We will define every element of $\mathbb{N}$ to be strictly smaller than $\infty$. Then $\mathbb{N}\cup\{\infty\}$ is well ordered, but your procedure, starting from $0$ will never reach $\infty$. In particular $\infty$ is not a successor of anything. And well ordered reals will have a lot such numbers, which are not successors of anything.