Since $H_0(\mathbb{D}^{n},\mathbb{S}^{n-1}) = 0$ I was wondering wheter there was a simple generalization of the fact that $H_0(X,A) = 0$. It should be true if $X$ is path connected with $ \varnothing \ne A \subseteq X$ ? $H_0(X,A) = C_0(X)/i_{\sharp}C_0(A)$ so if $H_0(X)$ is generated by points $H_0(X,A)$ is generated by the "costant" which doesn't belong to $A$.
If I take $\gamma$ with $\gamma(0) = x_0 \in X \backslash A, \gamma(1) = y \in A$ if I think to $\gamma$ as $1-$simplex with $\partial_1 \gamma = [C_{x_0}] - [C_y] = [C_{x_{0}}]$, then $\partial_1$ is surjective so every point is homologous to $[0]$.
Is this Correct ? In any case, any type of comment about this proof of or giving a real proof of this fact would be appreciated.
Your proof is correct. It uses the construction of singular homology groups and has thus a "geometric" flavor.
If you know the long exact homology sequence for reduced homology, then it is very easy to see. It ends with
$$ \dots \to \tilde H_0(X) \to H_0(X,A) \to 0$$
For path-connected $X$ we have $\tilde H_0(X) = 0$ and therefore $H_0(X,A) = 0$.