I have the following proof that I am trying to understand. I understand every part except why $A \cup \{v\}$ is an acyclic set. I get that $A$ is an acyclic set, but what guarantees that introducing $v$ does not introduce a directed cycle?

I have the following proof that I am trying to understand. I understand every part except why $A \cup \{v\}$ is an acyclic set. I get that $A$ is an acyclic set, but what guarantees that introducing $v$ does not introduce a directed cycle?

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Introducing $v$ cannot create a cycle because all edges between $v$ and $A$ are directed in the same direction: we chose $A$ from the subset $N^{-}(v)$.
A cycle that contains $v$ would have to use an edge oriented from $A$ to $v$ and an edge oriented from $v$ to $A$, which do not both exist. A cycle that does not contain $v$ - well, that goes back to the induction hypothesis, since that would just be a cycle in $A$.