[proofplan] The construction already gives the infinitude of $\mathbb{N} \setminus A$, so the only point is to rule out an infinite computably enumerable subset of the complement. Suppose such a subset exists and name it as some $W_e$ in the fixed enumeration of computably enumerable sets. The requirement attached to that same index forces $W_e$ to meet $A$, contradicting the assumption that $W_e$ is contained in $\mathbb{N} \setminus A$. [/proofplan] [step:Use the construction to keep the complement infinite] By the stated property of the simple-set construction, $\mathbb{N} \setminus A$ is infinite. Thus, to prove that $\mathbb{N} \setminus A$ is immune, it remains only to prove that it contains no infinite computably enumerable subset. [/step] [step:Assume an infinite computably enumerable subset of the complement exists] Suppose, for contradiction, that there exists a set $B \subset \mathbb{N}$ such that $B$ is computably enumerable, $B$ is infinite, and \begin{align*} B \subset \mathbb{N} \setminus A. \end{align*} Because $(W_e)_{e \in \mathbb{N}}$ is an effective enumeration of all computably enumerable subsets of $\mathbb{N}$, there exists an index $e \in \mathbb{N}$ such that \begin{align*} B = W_e. \end{align*} [guided] We argue by contradiction because immunity is a universal negative statement: no infinite computably enumerable set may sit inside the complement. So assume there is one. Let $B \subset \mathbb{N}$ denote such a set, with the three relevant properties: \begin{align*} B \text{ is computably enumerable}, \qquad B \text{ is infinite}, \qquad B \subset \mathbb{N} \setminus A. \end{align*} The fixed sequence $(W_e)_{e \in \mathbb{N}}$ lists every computably enumerable subset of $\mathbb{N}$. Since $B$ is computably enumerable, this means $B$ appears somewhere in the list. Therefore there is an index $e \in \mathbb{N}$ such that \begin{align*} B = W_e. \end{align*} This index is the one whose requirement in the simple-set construction will be used. [/guided] [/step] [step:Apply the requirement for the chosen index] Since $B$ is infinite and $B = W_e$, the set $W_e$ is infinite. Therefore the construction property for index $e$ gives \begin{align*} W_e \cap A \neq \varnothing. \end{align*} Choose $n \in W_e \cap A$. Since $B = W_e$, we have $n \in B$. Since $B \subset \mathbb{N} \setminus A$, this implies $n \in \mathbb{N} \setminus A$, hence $n \notin A$. But $n \in W_e \cap A$ also gives $n \in A$, a contradiction. [/step] [step:Conclude that the complement is immune] The contradiction shows that no infinite computably enumerable set $B \subset \mathbb{N}$ can satisfy \begin{align*} B \subset \mathbb{N} \setminus A. \end{align*} Together with the infinitude of $\mathbb{N} \setminus A$, this proves that $\mathbb{N} \setminus A$ is immune. [/step]