Let $G$ be infinite group and $G$ act on compact metric space $(X, d)$, $\varphi:G\times X\rightarrow X$.
$\varphi:G\times X\rightarrow X$ is called minimal action, whenever there is not proper closed set $A\subseteq X$ with $GA\subseteq A$. ($GA=\{\varphi(g, a)| g\in G, a\in A\}$).
Question. Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$. If $\varphi:G\times X\rightarrow X$ is minimal action. $\varphi|H:H\times X\rightarrow X$ is minimal action?
I'll give an example where the action $G$ is both minimal and free (admits no periodic points).
Let $X = S^1 \sqcup S^1 = \{(\theta,0),(\theta,1)\mid \theta\in S^1\}$ be the disjoint union of two circles and let $\rho\colon \mathbb{Z}\times S^1\to S^1$ be rotation by some irrational angle $\alpha$: $\rho(\theta)=[\theta+\alpha]$.
We extend this action to $X$ by defining the new action $\tilde{\rho}\colon \mathbb{Z}\times X \to X$ by $$\tilde{\rho}((\theta, i)=(\rho(\theta),i+1)$$ with addition being mod $2$.
As $\rho$ is minimal, so is $\tilde{\rho}$ (prove this). The subgroup $H=2\mathbb{Z}$ is not minimal, as each of the disjoint components of $X$ are fixed setwise by every element in $H$.
[edit] oops I missed that $X$ was meant to be connected. I'll try to find a new connected example.