Theorem
Let $D$ be an infinite subset of $\mathbb{Z^+}$. Then $D$ is denumerable, and in fact there is a unique enumeration of $D$, namely ${ k_1 , k_2 ,....}$ such that
$$k_1 < k_2 < ... < k_n < k_{n+1} < ....$$
Proof.
We let $k_1$ be the smallest element of $D$. Suppose inductively that we have defined $k_l < k_2 < ... < k_n $ in such a way that any element $k$ in $D$ which is not equal to $k_1, ... ,k$ is $> k_n$. We define $k_{n+ 1}$ to be the smallest element of $D$ which is $> k_n$. Then the map $n\rightarrow k_n$ is the desired enumeration of $D$.
Confusion
The proof states that $k_{n+1}$ is the smallest element in $D$ (....We define $k_{n+ 1}$ to be the smallest element of $D$...). Then it declares $k_{n+1}>k_n$ (...smallest element of $D$ which is $> k_n$.). Isn't this contradictory?