Stuck trying to prove conservation of energy
Maybe I'm not seeing the problem, I will gladly delete this if that is the case, but you do not need to assume anything additional. If you use your Eq. (3) this is what you get
$$ \frac{{\rm d}}{{\rm d}t}(T + U) = m \ddot{x}_k \dot{x_k} + \frac{\partial U}{\partial x_k}\dot{x}_k = \left[ m\ddot{x}_k - m\ddot{x}_k \right]\dot{x}_k = 0 $$
So Eq. (4) is just a trivial $0 = 0$, which is completely consistent
Unfortunately, you've made a fatal error at the very first step! It's wrong to interchange the $d/dt$ and $\partial / \partial \dot{x}_i$. This quickly leads to absurdities, such as in this question where the OP concludes the Euler-Lagrange equation is trivial. This point isn't explained well in most textbooks; for a related question see here. I'll explain why briefly, then show a derivation like yours that works.
In your setup, $T$ is a function of the two independent variables $q$ and $\dot{q}$, with no explicit dependence on time. To take a time derivative, you must restrict to a specific path $$q(t) = \gamma(t), \quad \dot{q}(t) = \frac{d\gamma(t)}{dt}.$$ In other words, the time derivative really means $$\frac{dT}{dt} \equiv \frac{d}{dt} T\left( \gamma(t), \frac{d\gamma(t)}{dt} \right).$$ Then $dT/dt$ is solely a function of $t$ and the path $\gamma$. It is no longer a function of $q$ and $\dot{q}$, so taking the partial derivative with respect to $\dot{q}$ doesn't even make any sense. The manipulations leading up to your equation (4) similarly don't make sense.
A big part of the confusion is that usually the path $\gamma(t)$ is written as $q(t)$, so a function of a path $q(t)$ looks unfortunately similar to a function of the variables $q$ and $\dot{q}$. Compounding the confusion, the Lagrangian can also have direct dependence on $t$ (in addition to $q$ and $\dot{q}$), but this still isn't the same as depending on the path $q(t)$. It's a big mess that's confused many students.
Here's a different way to do it, which stays true to the spirit of your derivation. By the chain rule, $$\frac{dE}{dt} = \frac{\partial T}{\partial \dot{q}} \frac{d \dot{q}}{dt} + \frac{\partial U}{\partial q} \frac{dq}{dt}.$$ Now, applying the equation of motion $$\frac{d}{dt} \frac{\partial T}{\partial \dot{q}} = - \frac{\partial U}{\partial q}$$ we have $$\frac{dE}{dt} = \ddot{q} \frac{\partial T}{\partial \dot{q}} - \dot{q} \frac{d}{dt} \frac{\partial T}{\partial \dot{q}}.$$ However, because $T$ is only a function of $\dot{q}$, so is $\partial T / \partial \dot{q}$, so by the chain rule again, $$\frac{dE}{dt} = \ddot{q} \frac{\partial T}{\partial \dot{q}} - \dot{q} \frac{d\dot{q}}{dt} \frac{\partial}{\partial \dot{q}} \frac{\partial T}{\partial \dot{q}} = \ddot{q} \left(1 - \dot{q} \frac{\partial}{\partial \dot{q}} \right) \frac{\partial T}{\partial \dot{q}}.$$ However, because $T$ is quadratic in $\dot{q}$, $\partial T / \partial \dot{q}$ is linear in $\dot{q}$. That means that differentiating with respect to $\dot{q}$ and then multiplying by $\dot{q}$ again does nothing, so $dE/dt = 0$ as desired.
Of course, as the existing answer says, it would be much faster to get this result if we just used the fact that $T$ was quadratic from the beginning, but I figured you wanted a derivation that used some of the partial derivative machinery.