Theorem 8. Suppose $P$ is an $n\times n$ invertible matrix over $F.$ Let $V$ be an $n$-dimensional vector space over $F,$ and let $\scr B$ be an ordered basis of $V.$ Then there is a unique ordered basis $\scr \overline{B}$ of $V$ such that
\begin{align} \tag{1}[\alpha]_{\scr B} = P[\alpha]_{\scr \overline{B}} \end{align} for every vector $\alpha \in V$
$\Big($ $[\alpha]_{\scr B}$ is the coordinate matrix of $\alpha$ relative to the ordered basis $\scr B.$ $\Big)$
Proof : Let $\scr B$ consist of the vectors $\alpha_1,\dots,\alpha_n.$ If $\scr \overline{B}= \{\overline{\alpha_1},\dots ,\overline{\alpha_n}\}$ is an $\color{red}{\text{ordered basis}}$ of $V$ for which $(1)$ is valid, it is clear that \begin{align} \overline{\alpha_j}=\sum_{i=1}^n P_{ij}\alpha_i. \end{align} Thus we need only show that the vectors $\overline{\alpha_j},$ defined by these equations form a basis.
and the proof goes . . .
I don't understand this argument given by the author wherein they assume the existence of $\scr \overline{B}$ as a basis and then finally prove that it is indeed a basis. I have a feeling that the highlighted colored text should be replaced as "ordered set". What is going on here?
They don't assume a basis but say that if it was a basis then equivalently a property, namely $ [\alpha]_{\scr B} = P[\alpha]_{\scr \overline{B}} $, will hold.
(Note that if the property holds, the set of coordinates of the vectors in $\scr\overline{B}$ include each coordinate of the vectors in $\scr B $ in their span, following which so will the corresponding vectors. So $\scr\overline{B}$ is indeed spanning and since it has the same number of elements as $\scr B $ is a basis. Similarly the converse uses the fact that there can only be a unique way of expressing the old basis in terms of the new one.)
They then proceed to prove that property.