For reference, Axler's proof was copied in full in this question. In the beginning of that proof, Axler chooses the nonzero vector $v$ without loss of generality. This made me doubt his proof because it is obvious that not all nonzero vectors are necessarily eigenvectors. So I looked for something that would help me to distinguish $v$ from other arbitrary vectors, and I realized that for it to be true that $(v,Tv,\dots,T^nv)$ is linearly dependent, each $T^i$ for $0<i\leq n$ must be nonzero. Thus I am left thinking that Axler's proof must show that such a $v$ can definitely be chosen. Was this step left out of the proof because the existence of such a $v$ trivial in some manner that I am not noticing? Or is my reasoning about the proof incorrect?
Well, the proof doesn't claim that $v$ is an eigenvector, so there's no problem if it isn't.
The vectors $(v,Tv,\dots,T^nv)$ are linearly dependent regardless of the values of $v$ and $T$. It doesn't matter whether $v$ is nonzero. It doesn't matter whether $T^i$ is nonzero. It doesn't even matter that $T$ is a linear operator. Any set of $n+1$ vectors is linearly dependent.