 Solution

On $S_{0,3}$, a simple proper arc joining two distinct specified punctures is unique up to isotopy relative to its ends. Indeed, a small regular neighborhood of the arc and its two ends has one boundary curve separating those two punctures from the third; the Jordan curve theorem gives the unique such separation, and the disc it bounds gives the isotopy between any two choices.

Choose the three joining arcs, one for each puncture pair, with disjoint interiors. They form an ideal triangle graph whose complement consists of two discs. A pure homeomorphism fixes all three endpoints and carries each arc to an isotopic arc. The simultaneous-isotopy lemma makes it fix the three arcs, and the Alexander trick on each complementary disc makes it isotopic to the identity. Therefore