I read in On the spectral gap for finitely-generated subgroups of SU(2) that every finitely generated dense subgroup of $ SU(2) $ contains a free subgroup.
Is it true in general that every finitely generated dense subgroup of $ SU(n) $ contains a free subgroup?
This follows easily from the Tits alternative, as pointed out by Moishe Kohan.
Let $ G $ be a connected Lie group that is not solvable. Let $ \Gamma $ be a finitely generated dense subgroup of $ G $. Since $ G $ is not solvable and $ \Gamma $ is dense in $ G $ then $ \Gamma $ cannot be solvable or virtually solvable (a finite index subgroup of a dense group is still dense). So by the Tits alternative $ \Gamma $ must contain a nonabelian free group.