Prove that $BPP^{BPP} = BPP$
Here's a sketch: you seem to be on the right track. Let $L\in BPP^{BPP}$ be decided by a BPP machine $M$ that makes queries to an oracle $A\in BPP$ decided by a $BPP$ machine $M'$. Let $E$ be the error for $M$, $E'$ the error for $M'$ (that is, the probability that the BPP machines fail is at most $E, E'$). We will choose $E, E'$ appropriately via error reduction. 

Now, fix a string $z$ to use as randomness for all queries to $M'$. There are at most polynomially many queries to $M'$, say at most $|x|^k$, since $M$ is a ppt TM. Thus, $$\operatorname{Pr}[\text{an error occurs in some query to } M']\le |x|^k\cdot E'$$ So the overall probability of error by $M$ in deciding $L$ is at most $|x|^k\cdot E' + E$, so if we choose $E=\frac{1}{6}$ and $E'=\frac{1}{6|x|^k}$ (which we may do by the standard BPP error reduction), we get an overall probability of error $\le\frac{1}{3}$, so the procedure described is a $BPP^P\subseteq BPP$ procedure.

By the same proof we have the more general statement that $BPP^{BPP^A}\subseteq BPP^A$.