TITLE: Well-ordered set countability
QUESTION [6 upvotes]: I know there are well-ordered sets that are not countable.
Suppose you are given an uncountable, well-ordered set $S$.
Isn't it possible to provide a bijection $f:\mathbb{N} \rightarrow S$ as following?
$S$ is well-ordered, so it has the smallest element, say $s_1$. $S \setminus$ {$s_1$} is also well-ordered, so there is the next smallest element, $s_2$, Similarly, there is the next smallest element $s_3$, and so on.
Continuing like this, define $f(i) = s_i$. 
I know there is something wrong with this, but I cannot really see why...

REPLY [4 votes]: Even if you made a mistake about surjectivity, as pointed out by 5xum and alex, you did a good thing: you proved that 
given any two well-ordered sets, one is always the intial segment of the other! (which is, of course, true. The proof needs transfinite induction in general, but works exactly as your attempt of building the bijection)