To prove the assert I led back to prove that $H_c^0(X) = 0$ thanks to Poincarè-duality, using the fact that $H_c^0(X) = \text{lim}H^0(X,X\setminus K,\mathbb{Z}) = \text{lim} 0 = 0$, since here we noticed that $H_0(X,A)= 0$ if $X$ is path connected for every nonempty $A$ (and using universal coefficient theorem for cohomology).
Is the reasoning is correct, one could imply something on $H_c^0(X,G)$ with $G \ne \mathbb{Z}$ or are there counterexample to the fact that $H_n(X,G) \ne 0$ with above hypothesis ?
You can prove directly that $H^0_c(X;R)=0$ where $R$ is any commutative ring (or an abelian group). The easiest way to see this is to observe that $H^*_c(X;R)$ can be described as the group of compactly supported cocycles (modulo compactly supported coboundaries). See for instance Hatcher's "Algebraic Topology," section 3.3. No need for direct limits.
In degree zero, coboundaries are irrelevant. If $X$ is path-connected then every compactly-supported degree zero cocycle is zero. Hence, $H^0_c(X;R)=0$.
Lastly, Poincare Duality works with coefficients in any commutative ring, again, see for instance Hatcher's book.