If I take a fixed $t \in R$, and $x$ were periodic, then $x(t) = x(t+T)$ and $E\circ x (t+T) - E\circ x (t) = 0$ would be given.
Using the fundamental theorem of calculus: $$E\circ x (t+T) - E\circ x (t) = \int_t^{t+T} {\nabla E(x(t'))\bullet x'(t') \; dt'} = -\int_t^{t+T} {||\nabla E(x(t'))||^2\; dt'}$$ the last expression doesn't equal $0$, unless $E(x) = const.$, which is in opposition to the periodicity of $x$.
Is this proof correct? I'm especially unsure about the correctness of the last statement. If not, how could I verify the assumption correctly?
Hint:
If $\dot{\mathbf{x}}=-\nabla E(\mathbf{x})$, then $g(t):=-E(\mathbf{x}(t))$ is monotone non increasing since $g'(t)=-\|\nabla E(\mathbf{x}(t))\|^2\leq 0$. Thus, the system cannot admit (non constant) periodic solutions. (Convince yourself that a real valued periodic function $g$ can not be monotone in the whole real line $\mathbb{R}$)