Let $G$ be a group which acts on the set $X$. Consider a subgroup $H$ of $G$ which acts on $X$ by the restriction of the action of $G$ on $X$. Let $[H\backslash G]:=\{g_i\ \ :\ \ \exists!\mathcal{O}\in H\backslash G \text{ such that }g_i\in\mathcal{O}\}$ be a set of representers of the $H$-orbits (i.e. a fundamental domain of the action by translation of $H$ on $G$) and let $\mathcal{D}$ be a fundamental domain of the action of $G$ on $X$. Let $\mathcal{D}_H$ be defined as: $$ \mathcal{D}_H:=\bigcup_{g_i\in[H\backslash G]} g_i\cdot \mathcal{D} $$ Show that $\mathcal{D}_H$ is a fundamental domain of the action of $H$ on $X$.
My work:
Let $x\in X$. Then $\exists!\mathcal{O}\in G\backslash X$ with $x\in\mathcal{O}$. Since $\mathcal{D}$ is a fundamental domain, $\exists!d\in\mathcal{D}$ with $\mathcal{O}=G\cdot d$. The there exists $g\in G$ with $x=g\cdot d$. Furthermore $\exists!\mathcal{O}_H\in H\backslash G$ such that $g\in\mathcal{O}_H$. Thus, $\exists!g_i\in[H\backslash G]$ with $\mathcal{O}_H=H\cdot g_i$ and thus there exists an $h\in H$ with $g=h\cdot g_i$. So we see that $x=h\cdot g_i\cdot d$ wich implies $x\in H\cdot\mathcal{D}_H$. So we have shown $X=H\cdot\mathcal{D}_H$.
However, I fail to see why $\mathcal{D}_H$ is a fundamental domain, i.e. $H\cdot d_H=H\cdot d_H'\implies d_H=d_H'\ \ \forall d_H,d_H'\in\mathcal{D}_H$.
My progress here:
Assume that for $d_H,d_H'\in\mathcal{D}_H$ we have $H\cdot d_H=H\cdot d_H'$ and thus $d_H=hd_H'$ for some $h\in H$. By construction of $\mathcal{D}_H$ we have $d_H=g_id$ and $d_H'=g_jd'$ for some $g_i,g_j\in\left[H\backslash G\right]$ and $d,d'\in\mathcal{D}$. Thus $g_id=hg_jd'\implies Gg_id=Ghg_jd'\implies Gd=Gd'\implies d=d'$ since $d,d'\in\mathcal{D}$. Therefore $d_H=g_id$ and $d_H'=g_jd$ which leads to $Hg_id=Hg_jd$. If this would imply $Hg_i=Hg_j$ then we would be done since $\left[H\backslash G\right]$ is a fundamental domain. Is this implication true, and if yes how to prove it?
I still don't see how to prove nor how to disprove it. Any help would be great!