About minimal group actions? 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? 
 A: Let us take your question true i.e., by the assumption on $G$, $X$, $H$ and $G\curvearrowright X$, we have that $H\curvearrowright X$ is minimal. I assume further assumption that $H$ is a normal subgroup of $G$. In this case, $G/H\curvearrowright X/H$ minimally which by $X/H$, I mean the orbit space of the action of $H$ on $X$, endowed with quotient topology by the map $X \rightarrow X/H$, which is connected space. But $G/H$ is a finite group which acts minimally on $X/H$, therefore, $X/H$ is a singletone that means the action of $H$ on $X$ is point transitive i.e., has only one orbit.
Let $*$ denote the sentence "Suppose $(X,d)$ is a connected compact metric space and $H$ is a subgroup of finite index in group $G$ and $G\curvearrowright X$ minimally". 
Now we must have a true statement: $* \Rightarrow H\curvearrowright X$ is point transitive. But this statement is false (remember irrational rotation of integers $\mathbb{Z}$ on $S^1$). 
So by a logic argument, we deduce that your question is false.
A: 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.
