Procedure for producing a total order on infinite set

Question

Suppose we want to prove that every infinite set, $X$, can be totally ordered. A probably faulty procedure I thought of was the following: by the Axiom of Choice we know that there exists a surjection $f:X\to\mathbb{N}$. If the preimage $f^{-1}(n)$ contains one element we say $f^{-1}(n)<x\in f^{-1}(m)$ if $n<m$. If $f^{-1}(n)$ is finite we can order this preimage set by using the bijection with $\{1,2,\dots,|f^{-1}(n)|\}$. If $f^{-1}(n)$ is infinite we apply this procedure again using the surjection $f':f^{-1}(n)\to\mathbb{N}$. It might be that this does not stop in finite steps. So should it stop, for constructing a total order? If so, why?