Let $G$ be an algebraic group and $H \subset G$ a subgroup of $G$. Let $H_0$ be a subgroup of $H$ of finite index. Then I guess the Zariski closure of $H_0$ is exactly the Zariski closure of $H$ in $G$, since intuitively it should be.
But I have no clue how to prove it, anything will be helpful.
Note that this is equivalent to asking whether any finite index subgroup of $H$ is dense. In particular there's no real reason to consider the situation within an ambient algebraic group $G$.
If you assume irreducibility, or even connectedness, of $H$, the claim is true as the only finite index closed subgroup of $H$ is $H$ itself (This follows as $H$ is the union of the cosets of any such subgroup), yet the closure of any finite index subgroup will be a finite index subgroup as well.
Without assuming connectedness of $H$ the claim is not true. Consider for example $H_0\subsetneq H\subsetneq G$ finite groups.