Equivalent definition of compact support modulo $H$ 
My question is how to prove the equivalenc:

Given a function $f\in X$, the following definition for compact supported modilo H are equivalent:
  
  
*
  
*The image of the support supp $f$ of $f$ in $H\backslash G$ is compact.
  
*supp $f\subset HC$ for some compact set $C$ in $G$.

I found a similar question here that may help(link), but after reading that I still can't solve it myself.
 A: It's been a long time, but here's an answer.
Denote $q : G \rightarrow H \backslash G$ the quotient map.


*

*$1$ implies $2$: There exists an open compact subgroup
$K \le G$ such that $f \left( g k \right) = f \left( g \right)$ for
any $g \in G$ and $k \in K$. Then since $q$ is an open map, we have that $q\left(
   K \right)$ is open. Now $q \left( \operatorname{supp}f \right)$ is
compact, and we have $$ q \left( \operatorname{supp}f \right)
   \subseteq \bigcup_{g \in \operatorname{supp} f} q\left( g K
   \right).$$ Since $q \left( \operatorname{supp}f \right)$ is compact,
we have that there exists a finite subcover, i.e., there exist $g_1,
   \dots, g_N \in G$, such that $$ q \left( \operatorname{supp}f \right)
   \subseteq \bigcup_{i = 1}^N q\left( g_i K \right).$$ This implies
$$ \operatorname{supp}f \subseteq H \bigcup_{i=1}^N g_i K,$$
and $C = \bigcup_{i=1}^N g_i K$ is compact as a finite union of
compact sets.

*$2$ implies $1$: suppose that $\operatorname{supp}f \subseteq H \cdot C$, where $C \subseteq G$ is a compact subset. It suffices to explain why $q\left( \operatorname{supp}f \right)$ is closed, because then $q\left( \operatorname{supp}f \right) \subseteq q\left( HC \right) = q\left( C \right)$, and $q \left( C \right)$ is compact as a continuous image of a compact set. Suppose $g_0 \in H \backslash G$ and $g_0 \notin q \left( \operatorname{supp}f \right) $. Let $g \in G$ with $q \left( g \right) = g_0$. Then $f \left( g \right) = 0$. Since $f \left( g k \right) = f\left( g \right)$ for any $k \in K$, we have $f\left(gk \right) = 0$. Hence $q\left( g K \right) \cap q \left( \operatorname{supp}f \right) = \emptyset$. We have that $q\left( g K \right)$ is open, since $q$ is an open map, and we have $g_0 \in q\left( g K \right)$. Therefore we showed that the complement of $q \left( \operatorname{supp} f \right)$ in $H \backslash G$ is open, i.e., $q \left( \operatorname{supp} f \right)$ is closed. Since $q \left( \operatorname{supp} f \right) \subseteq q \left( C \right)$, where the left hand side is closed and the right side is compact, we get that the left side is compact. 

