Is this a counterexample to ΔU = 0 J for isothermal processes? In class, I have learned that for isothermal processes the following is true at all times.

$$\Delta U=q+w=0\text{ J}$$

But suppose I lift an object against the force of gravity. Then I (the surroundings) have done work on the object (the system) without exchanging heat (neglecting friction).

Now, I claim that because some nonzero work has been done on the system, and no heat has been exchanged, the change in internal energy is nonzero:

$$\Delta U=w\neq0\text{ J}$$

Are my assumptions about internal energy incorrect or is there something totally wrong about my approach?
The situation you cite calls for the use of the more general form of the first law of thermodynamics (which I assume your textbook provides), given by $$\Delta U+\Delta (KE)+\Delta (PE)=q+w$$where KE is the organized kinetic energy of the system and PE is the potential energy.  Applying this to the problem at hand gives $$\Delta (PE)=w$$and the change in internal energy is zero.  So the thing that was wrong with your approach was not using the proper form of the first law.