Does there exist a non empty perfect subset of $R$ containing only transcendental numbers ? I am looking for an example although even an existential proof will suffice.
Here's a non-constructive proof. I'm still thinking about a constructive one.
The Cantor set is homeomorphic to its own square (i.e. $C \simeq C \times C$), and from this it follows that $C$ contains infinitely many disjoint copies of itself. If we construct $C$ on an interval with transcendental endpoints, then since the algebraic numbers are countable, then at least one of these copies of $C$ cannot contain any algebraic numbers.
I think a constructive proof using Cantor sets is also possible, but I'm not 100% sure. If I figure something out, I'll update.