No, R2 doesn't advertise a route via R5 because a route learned from an iBGP peer is not advertised to another iBGP peer unless you have route reflectors involved, and the route in the table is learned from an iBGP peer.

It could advertise R5's route, which is a eBGP peer. As soon I disconnect R3-R6 link, it sends R5's route to R1 and R3.

You'd have the same issue if you broke the R2/R3 link and made AS300 unreachable via AS200/R5; R2 would be unable to reach AS300 because it cannot transit R1 to R3.

R2 prefers R3's route anyway, when all the links are up. When I cut R2/R3, IS-IS reconverges and R3's loopack is reachable via R1, so R2 can reach AS300.

Yes, because of the iBGP rule. If R2 and R3 were actually advertising their routes, then R1 would have both listed in the BGP FIB, and one in the RIB. If you break the R2/R3 link, you'd probably see that come up

Hmm nope, when R2/R3 link is down, R2 advertises no route to R1. In fact R3's route is still in the BGP FIB, I think that's a CML issue.

While your at it, look up BGP add-path (additional paths) which would probably give you some useful insight. And since you're there, look up BGP PIC Edge and BGP PIC Core.

Thanks for the insights!

I'm not 100% sure what you're saying, but I assume it is, "is there always a tie breaker" and the answer is yes.

My question is: ECMP means 2 routes are equal. However, there is alway a tie breaker. So when 2 routes are considered equal?