I'm having trouble understanding why we make the assumption in the highlighted part of the proof, any help would be appreciate, thanks.
Proposition. $\forall G = (V,E)$ undirected graph, the pair $(E, \mathscr {F}_G)$ is a matroid. $$\mathscr {F}_G = \{A \subseteq E: (V,A)\space \text{is a forest}\}$$ Proof uses this lemma
Lemma. Every forest with $n$ nodes and $k$ trees has $n-k$ edges.
Proof. Consider $A,B\in \mathscr {F}_G$, $|B| = |A| +1 $ and $|V| = n$, than $(V,A)$ and $(V,B)$ has the same number of nodes $n$ but $(V,A)$ has an edge less and so one more tree.
$\exists u,v \in V $ in the same tree in $(V,B)$ but different tree in $(V,A)$, so exist a path $C = ( u=b_0,b_1, ...,b_n=v)$ connecting $u\to v$ in $B$. Must exist an edge $(b_i,b_{i+1}) \notin A$ and in two different trees in $A$.
So $$A\cup\{(b_1, b_{i+1})\}$$ does not form cycles and $(V, A\cup\{(b_1, b_{i+1})\})$ is a forest. $\blacksquare$
Why can't we just suppose the existence of $b_i,b_{i+1}$ call them $u,v$ without mentioning $C$, is less precise or is it wrong? To me it doesn't seem to be a problem but if the proof is written as it is surely there must be a reason.