Proof: (by contradiction)
Let $\langle B_i\rangle=P_i$; the set $B_i$ span $P_i$ such that $P_i$ is the set of all polynomials of degree $i$. Suppose there exist a set $S$ such that $S:=\{p_i(x)\in B_i \mid p_i(x)\in\bigcup_{i=1}^n B_i \text{ where $B_i$ is the basis of } P_i\}$ and $\langle S\rangle=P$. This implies that $\forall p_j(x)\in P$, $p_j(x)=\sum_{i=1}^n c_ip_i(x)$ for some $p_i(x)\in B_i$ and $c_i\in\Bbb{F}$. However, if $j\geq n+1$ then $p_j(x)\notin \langle\bigcup_{i=1}^n B_i \rangle$. Thus, a contradiction. $\Bbb{QED}$
You have not given any proof that no finite set spans $P$. Your argument is very difficult to understand, but it appears you have argued that one particular finite set $S$ does not span $P$. That is not enough: you need to start with some completely arbitrary finite set $S$ which spans $P$ and get a contradiction. You cannot assume $S$ has any specific definition as you have done.