you seem to have a mistake, you dotted one side of the 3rd equation with r dot and the other with r double dot

is not $E_0$ in your equation simply $m|\dot{r}(0)|^2/2$, that is integration constant? Then it must be constant.

the alternative way of characterizing conservative forces is that for such forces you can defined function of position only, which will be conserved with kinetic energy. That is, you want to show, that integral $\int_0^T|\dot{r}(t)|dt$ cannot be rewritten as function of position only, but it necessarily depends on the trajectory.

Thank you @Wolphramjonny, fixed it.

@Umaxo Yes, I guess that is true. So how should I argue in the end?

@samwolfe how about doing integration for closed trajectories? If $r(0)=r(T)$, then for conservative forces you also get $\dot{r}(0)=\dot{r}(T)$, because the kinetic energy should return to original value. But this is impossible in your case, because the integrand is always positive, and therefore kinetic energy after one (or more) loop will be necessarily less than the original value

what about showing that in a closed trajectory the total work is not zero?

@Wolphramjonny please check the note at the end of the question.

Oh sorry, i missed it!

@Umaxo, I see your point, but I still don't get what is wrong with my proof. Could you please expand?

@samwolfe you showed that kinetic energy is dissipated for $T\rightarrow \infty$. How does it imply that force is not conservative? You just said so, but I don't think it is so simple. Conservative force is force for which no (position dependent only) potential can be defined. I think you should show that dissipation of this kinetic energy imply impossibility of potential, not just say so. Naively, one could think that the kinetic energy is just transferred to potential energy, so naively there is no problem with kinetic energy tending to zero as time grows.

Exactly! That's what I'm missing. So how do I argue that we have no potential energy and thus energy is dissipated?

I think a much more useful thing would be to just introduce work. you've unknowingly used the concept of 'power' when you dot the velocity with the force. You could simplify all of these arguments with the idea of work. so why not?

I agree, I'll give you some context: I'm a TA for a Newtonian Mechanics course, and the lecturer wrote this on the board. Students don't know what work is, and I've been trying to understand his writings, which is essentially the question.

@samwolfe As I said, take a closed loop. The dissipation of kinetic energy does not depend on the trajectory. You showed this for T approaching infinity, but this will force you to make infinite loop and the whole thing gets polluted with limits. The dissipation of kinetic energy is, in your case, universal. It holds $E_k=E_0-U(T)$, where U is positive no matter the trajectory (it is value of U that depends on the trajectory, not its sign). Therefore after one closed loop the kinetic energy is necessarily smaller and you have two values of $E_k$ at single place => no potential can be defined

What if there are no loops?

@samwolfe You got me there. Obviously you can curve the path anyway you want by perpendicular forces (because they are not doing any work), but I don't see how this can be justified without the concept of work. Then how about comparing particles with different initial velocities that start and end at the same place and showing that $E_k-E_0$ is not a constant same for every particle (which would be the case if you could define potential which depends only on the position)?

@Wolphramjonny you can define potential energy as function of position of the particle which is conserved for motion of the particle when taken together with its kinetic energy.

@Umaxo, yes, that is right

@Umaxo how would that be mathematically expressed, in my case?

I give up, let's use work. How would that change my proof now?

Dear both, I believe I have found a solution. I've posted it as an answer.