It can be shown easily that there exists subgroup(s) such that $S$ is union of its left cosets. Trivial example: Take the trivial subgroup $H_0=\{e\}$ and then $S=\bigsqcup\limits_{g\in S}gH_0$.
Then I constructed a POSET, $$X=\{H\subseteq G\mid H\text{ is subgroup and }S\text{ is union of left cosets of }H\},$$
order $X$ by $H\le H'\iff H\subseteq H'$. I was thinking to apply Zorn's lemma here. But I'm unable to prove the hypothesis of Zorn's lemma (the condition "union of left cosets" is creating problem).
Then I made some observations such that if $H$ is a subgroup of $G$ such that $S=\bigsqcup\limits_{g\in I}gH$ where $I\subseteq G$ is indexing set. Then $I\subseteq S$ and $H\subseteq \bigcap\limits_{g\in I} g^{-1}S$.
This exercise is given under the section of Group Action, so I have tried to think using actions but got nothing useful.
Can anyone help me to solve the problem? Thanks for your help in advance.
First, let us remark that for a subgroup $H$, we have $S = \cup_{g\in S}~ gH$ if, and only if : $$HS = \{hg : h\in H, g\in S\} \subset S$$
Using this, it is not hard to show that the condition of Zorn's lemma are satisfies.
But we can actually construct the maximal subgroup explicitly, by considering the action of $G$ on itself on the left.
What we are actually looking for is the stabilizer of $S$ : $$H_{\max} = \operatorname{Stab}(S) = \{ h\in G\mid hS\subset S\}$$
Let me know if you want me to fill in more details in some part of the proof.