Thompson's group $T$ of the circle itself contains a copy of $F_2$: this is indeed easy to find a ping-pong pair starting from 4 disjoint intervals. That $T$ is non-amenable is even easier, because it does not preserve any probability on Borel subsets on the circle (immediate once we check that the only probabilities on Borel subsets of the circle, invariant by $F$, are supported by $\{0,1\}$).

Thompson's group $T$ of the circle itself contains a copy of $F_2$: this is indeed easy to find a ping-pong pair starting from 4 disjoint intervals. That $T$ is non-amenable is even easier, because it does not preserve any probability on Borel subsets on the circle (immediate once we check that the only probabilities on Borel subsets of the interval $[0,1]$ invariant by $F$, are supported by $\{0,1\}$, and hence after identifying $0=1$ the only $F$-invariant probability on the circle is the Dirac at $0=1$, which of course is not $T$-invariant).