But then someone talked to me about Principle of Equivalence and not possibly being able to identify what is proper acceleration and what is coordinate acceleration with an accelerometer. Is it true ?
That's not true. By definition, an ideal accelerometer measures proper acceleration.
It appears you (and possibly the acquaintance who talked to you) are mixing and matching concepts from Newtonian mechanics and general relativity. Don't do that! Inertial frames in Newtonian mechanics and general relativity are rather different beasts.
The concept of an inertial frame is extremely important in Newtonian mechanics, not so important in general relativity. Proper and coordinate acceleration are concepts from relativity theory. In Newtonian mechanics, gravitation is a real force, but accelerometers can't measure it.
Newton's first law conceptually provides a way to test whether a frame is inertial: Simply find a test particle on which the net force is zero. Does the object appear to obey or disobey Newton's first law? The only problem: Good luck with this search!
That approach works nicely in general relativity. In fact, Gravity Probe B used exactly this approach. An accelerometer that registers zero acceleration does make for a local inertial frame in GR. Gravity Probe B flew low, so it was subject to drag. It had a free-floating test mass in its core. It used its thrusters to force the main part of the probe to accelerate just so and keep that free-floating mass centered. In doing this, Gravity Probe B was flying inertially from the perspective of general relativity. Because it was flying low, it was subject to some of the finer aspects of general relativity, and scientists could thereby use the observed motion as a test of general relativity.
The conclusion: Work done is dependent on different frames.
Let's look at a simple example. Say there is an object(1kg) that is initially at rest and we apply a constant force to move it for a certain distance to reach 2m/s in the rest frame(call it O). The KE change in frame O is 2J. Now consider a moving frame O' moving with a velocity=1m/s to the left with respect to frame O. In frame O', the object is initially moving to the right with a speed of 1m/s(NOT at rest), and the final speed is 3m/s, the KE change in frame O' = 4J, which is different from that in O. We know that in this simple case the KE change must equal to the work done on the object, and this must be true in both frames. Thus, we conclude the work done is different in different frames. This is not difficult to understand as in frame O', the object is moving with a greater average speed when the force is acting on it. Since the time that the force applied on the object must be the same in both frames, the displacement of the object in frame O' is indeed larger and the resulting work done is also larger.
Best Answer
Of course the theorem is still valid! To see why, let's review what goes into the proof of the work-energy theorem, which I'll state as $$\Delta K = \int \mathbf{F} \cdot d \mathbf{x}, \quad K = \frac12 m v^2.$$ The easiest way to prove this is to work differentially, $$dK = \mathbf{F} \cdot d \mathbf{x}.$$ By the definition of the differential, $$dK = m \mathbf{v} \cdot d \mathbf{v} = m \frac{d\mathbf{x}}{dt} \cdot d \mathbf{v} = m \, d \mathbf{x} \cdot \frac{d \mathbf{v}}{dt} = m \mathbf{a} \cdot d \mathbf{x}.$$ Each of the steps here requires no physical input at all, besides the definitions of $K$ and $\mathbf{v}$, and some basic calculus like the product rule and the chain rule. So, we see the only physical assumption needed is $$\mathbf{F} = m \mathbf{a}.$$ This is of course true in an inertial frame.
Now recall why fictitious forces are used. Going from an inertial reference frame to a noninertial one changes the acceleration. Therefore, naively it makes $\mathbf{F} = m \mathbf{a}$ stop working. The whole point of introducing fictitious forces is to adjust $\mathbf{F}$ so that $\mathbf{F} = m \mathbf{a}$ is true again. Then, as long as this holds, the proof of the work-energy theorem goes through exactly as above, so the theorem holds in non-inertial reference frames if you count the work done by the fictitious forces.