I believe, if we have a closed integral curve in a vector field then it is non-conservative. The idea is that say if it were conservative then we have a potential function say $\phi(x)$. Which gradient will be the vector field $\nabla\phi(\bar{x})$. We also know that the gradient tangent to the curve (integral curve) points to the biggest increase/decrease of the function. But we know that the integral curve ends up at the same point since it is closed, then the $\phi$ function value should not have increased at all, this implies that $$\nabla\phi(\bar{x}) = 0$$ which is true only if the tangent vector for our parameterization is zero i.e. $\bar{x}(t) = 0$, here obviously it cannot be a closed curve. Which we also see in the following idea
$$\oint\limits_{\gamma} w = \int_{a}^{b} \frac{1}{{\lambda(t)}} \left(\bar{x}(t)\right)^2 dt = 0 \implies \bar{x}(t) = 0$$
Is this argument correct? How could I make it more rigorous as right now it feels more intuitive? Thanks for any suggestions or ideas!!
As stated, your belief is not quite true, since point curves $x(t) = x_0$ are closed integral curves for conservative vector fields with critical points, i.e. $F(x_0) = \nabla\phi(x_0)=0$. However, you can show that if $F$ is conservative then $\dot{x}(t)=0$ for any closed curve $x(t)$.
Indeed if $F =\nabla\phi$ then $\dot{x}(t) = F(x(t))$ implies $$|\dot{x}(t)|^2 = \nabla\phi(x(t))\cdot\dot{x}(t)=\frac{d}{dt} \phi(x(t))$$ by the chain rule. But if $x(0)=x(b)$ for some $b$ then by the fundamental theorem of calculus
$$\int_0^b |\dot{x}(t)|^2 dt = \phi(x(b))-\phi(x(0)) = 0$$
And since $|\dot{x}(t)|^2 \geq 0$, this implies $\dot{x}(t)=0$ for all $t$.