$UP=BPP$ implies $UP=RP$?
Suppose $L\in \mathsf{UP}$, and let $M(x,y)$ be a polynomial deterministic Turing machine such that:


$x\in L \Rightarrow \exists !y : M(x,y)$
$x\notin L\Rightarrow \forall y : \neg M(x,y)$


Consider the language $L'$ consisting of words in $L$ along with a prefix of a witness:

$L'=\left\{(x,w)\big|\exists w' : M(x,ww')\right\}$

Clearly $L'\in\mathsf{UP}$. Given an oracle for $L'$, one can find the witness for words in $L$ in polynomial time. If $\mathsf{UP\subseteq BPP}$, then $L'\in\mathsf{BPP}$, so you don't actually have an oracle for $L'$, but you have an oracle which is right with high probability.

Now, given $x$, execute the machine for finding a witness for $x$ with an $L'$ oracle, only that now you replace oracle calls with executions of the $\mathsf{BPP}$ machine (after some amplification, I leave the details for you). If the execution resulted in a valid witness $w$ for $x$, i.e. $M(x,w)$ outputs $1$, then output "yes", otherwise output "no". Clearly this solution can only be wrong for positive instances, which makes it a good candidate for an RP machine for $L$. I leave it to you to choose the right success probability to make everything work.

Essentially, the above solution uses the property of self reducibility, and in that sense the same proof as that for $SAT$, or any other self reducible problem in $\mathsf{NP}$.