I was wondering whether it would be possible for a bullet shot from gun to become moon's satellite. How fast the bullet would have to be?

Anything projected from the surface of a planet should work aginst the gravity to go up or away from the planet. If  v is  the velocity of bullet of mass m, projected on a surface of the planet (here, moon.), of mass M and radius R, then the object's (here bullet's)  kinetic energy and potential energy should exceed that of its potential energy. The kinetic and poential eenergy of the object  are given by:

KE = mv^2/2 and

PE = m*g, where g is the acceleration due to gravitational attraction, given by:

mg =  GMm/R^2. Here, G is the unineversal gravitational constant = 6.673*10-11 Nm/kg^2, by tables. Or

g = GM/R^2

To escape the moon's gravity, the object's (bullet' s) kinetic energy should be more than  of it's surfacial potential energy, for which the required condition is:

bullet's KE > = bullet's  PE  on the surface of the planet (moon) or

mv^2/2 >= mg, or

V^2  =  2g  or

V^2 > = 2 moon's surfacial acceleration due to moon's gravity.

= 2* (GM/R^2), or

v = (2GM)^(1/2) / R.

For the moon of us (earth), by tables, M = 7.353*10^22 kg, R = 1.738* 10^6 meters. Therefore the escape velocity for the bullet (or any other thing projected from the moon's surface) is given by:

v = [(2*6.672*10^-11 Nm/kg^2) * (7.353*10^22 kg)]^(1/2) /(1738*10^6 m)

=1.274509489 meter/ second.