Short answer: position momentum uncertainty exists because their operators do not commute. Likewise the time and energy operators do not commute.
First: Physics is an empirical science so "proof" must be an experiment. Sometimes thought experiments will be allowed because they are useful for building and testing models. Mathematical proofs are not physical proofs.
Synthesis in physics involves building models (usually mathematical models) to explain observed behavior. In QM these intellectual edifices have proceeded quite far.
Within the contexts of those math models for QM, I would note the following:
1) The "state" of a QM system is represented by a "wavefunction", \psi, whose amplitude gives information about the expected outcomes of experiments.
2) Various observations are seen as operators on wavefunctions. For example, position is the operator "x". Interestingly in building the models, momentum has been interpreted as "\partial / \partial \Vec(x)" - i.e. gradient. Why? well because it explains observed experiments. (again - the standard of "truth" or "proof" or derivation in physics)
3) Predicting measurements of observables is a functional acting on the wave-function and the operator. So for instance, the expectation of measuring position is <\psi|x|\psi> (using bra ket notation)
4) Position-momentum uncertainty relation: note the momentum operator, P=\partial/\partial \vec(x) does not commute with the position operator. I.e. Px <> xP in fact the anti-commutator [P,x] \defined Px - xP = 1
( when one notes that I have used units where h-bar = 1 )
Remember to consider P and x as operators and apply the chain rule for differentiation. This anti-commutation of the operators is, within the constructed models on QM, the source of the uncertainty relation. I.e. not all operators commute.
5) Now consider time and energy. In order to be observed, both must be cast as operators acting on wavefunctions. What are the operators for time and Energy? - simple. The time operator is just multiplication by t. The energy operator is E\ defined \partial / \partial t. --- See how the Energy-time uncertainty relation looks like the position momentum uncertainty relation?
Q.E.D. Well as close as Q.E.D. is meaningful for physics since physics<> math. The real crucible comes in how this prediction from the math model used for QM predicts and agrees with experiment. And it does- very well.
A. In this sense, we do not prove E-t uncertainty from x-P uncertainty but rather use the mathematical architecture developed for QM in general.
B. Time being a dynamical variable is and is not a problem. If you want to measure time, you must define an operator to do that. In that sense it is not different than measuring position or spin or charge or ...
C. Time as a "special" dynamical variable can be removed by taking a Langrangian point of view, but that is another discussion that leads down the Feynman path.
D. The E-t uncertainty relation drops out quite naturally from a relativistic approach. When 3-vectors become 4-vectors, you better have t-E uncertainty otherwise Heisenberg will be quite angry. So in that sense one could "derive" E-t uncertainty from x-P uncertainty by requiring good behavior under relativistic considerations, but in my mind, just noting that operators don't commute is a more basic approach. (Clearly a personal intellectual aesthetic judgment, but I think shared by many.)