Are there any examples of $H < G$ such that for any pmp ergodic action of the group $G$ on a standard proba space $(X,\mu)$ there exists a set $A$ of $\mu(A)>C$ such that the action of the subgroup $H$ on $A$ is ergodic? It seems that this can happen only in finite index case...
Conclusion: The combination of the answers of Alex and Clinton give the complete answer to my question.
We seem to be talking past each other in the limited space provided by the comments, so maybe I can express myself better in the room provided by the answer box. You indicated that you were focused on countable discrete groups. For countable discrete $G$ and $H < G$ the following are equivalent.
There exists a constant $c > 0$ such that for any measure-preserving ergodic action of $G$ on a standard probability space $(X,\mu)$ there is a measurable $H$-invariant subset $A \subseteq X$ on which $H$ acts ergodically and with $\mu(A) \geq c$.
$H$ has finite index in $G$.
(not 2) $\Rightarrow$ (not 1). If $H$ has infinite index, consider the ergodic action of $G$ by shifts on $[0,1]^{G/H}$ with the usual product measure. $H$ doesn't act ergodically on any set of positive measure, as the components of its ergodic decomposition are null. More concretely, if $A \subseteq [0,1]^{G/H}$ is an $H$-invariant set of positive measure, there are disjoint sets $B,C \subseteq [0,1]$ such that a positive measure of elements of $A$ send the coset $H$ to something in $B$, and the same for $C$. Since the $H$ action doesn't shift this coset, this shows $H$ doesn't act ergodically on any set of positive measure. (A better way of writing this argument is simply that $f \mapsto f(H)$ is a null-to-one $H$-invariant Borel function from $[0,1]^{G/H}$ to $[0,1]$, which is enough to preclude ergodicity of the $H$ action on any non-null set.) Thanks to Robin for fixing the error in the original answer.
2 $\Rightarrow$ 1. Say $H$ has index $n$ in $G$, and fix coset representatives $g_1, \ldots, g_n$. Fix a measure-preserving ergodic action of $G$ on $(X, \mu)$. Suppose that $B \subseteq X$ has positive measure and is $H$-invariant. Then $\bigcup_{i \leq n} (g_i \cdot B)$ is $G$-invariant and $\mu$-positive, so by ergodicity has measure $1$. This implies that $\mu(B) \geq 1/n$. So picking $A$ to be an $H$-invariant set of smallest positive measure, $H$ acts ergodically on $A$ and $\mu(A) \geq 1/n$. Thus $c = 1/n$ works for all actions.
You can take $G$ to be any simple Lie group with finite center, and $H$ to be any non-compact Lie subgroup. Then, if the action of $G$ is ergodic, so is the action of $H$. This statement is called the "Moore ergodicity theorem".
In fact, both the actions of $G$ and $H$ will be automatically mixing. This follows from the theory of unitary representations of $G$.
Edit: Misread the question.
Let $X = \lbrace 0,1 \rbrace^\mathbb{Z}$ with the $(\tfrac{1}{2},\tfrac{1}{2})$ Bernoulli measure and let $T$ be the shift map $(T\omega)(n) = \omega(n+1)$. Put $G = \mathbb{Z}^2$ and define a $G$-action $S$ on $X$ by $S^{(n,m)} = T^{n + m}$. Take $A = X$ and $H = \mathbb{Z} \times \lbrace 0 \rbrace$. Since $T$ is ergodic both $S$ and $S$ restricted to $H$ are ergodic.
