The orbit of a moon of a close-in planet becomes stable for an indefinite period of time within two scenarios.
In Scenario 1, the moon is captured or formed above synchronism. Orbiting away, like our Moon, it is simultaneously slowing down the planet's rotation rate. If the mutual tidal synchronisation takes place before the moon crosses the reduces Hill radius, the stable configuration is reached and the moons survives infinitely. For this to happen, the moon must be massive (or the planet must be small). The minimal ratio of their masses must obey
$$
\frac{M_m}{M_p} > \left(\frac{R_p}{r_H^{\,\prime}}\right)^2\,\;,
$$
$R_p$ and $r_H^{\,\prime}$ being the planet's radius and the reduced Hill radius (Makarov & Efroimsky 2023, eqn 52), also available in ArXiV.
In Scenario 2, the moon is captured or formed below synchronism. Spiraling downwards, it is reducing the rotation rate of the planet. If the mutual synchronism is attained before the moon crosses the Roche radius, the configuration becomes stable. In this case, again, the ratio $M_m/M_p$ should not be too small, see Section 9.2 in Ibid.
As an entertaining aside, I would mention that a planet synchronised by its moon will thereby be nonsynchronous with respect to its host star. Now recall that a rotation rate faster than the mean motion works to tidally increase the semimajor axis in the tidal two-body problem. This will prevent a close-in planet from falling on its star. So, if you ever come across a planet, which is very close to its star but is not tidally descending, it may be right to enquire if that planet has a synchronous moon braking its tidal descent. Another possible feature of such a configuration would be the planet's finite eccentricity. (Fast rotators tend to boost $e$.)
