Why is every homogeneous space with $G$ action isomorphic to $G/H$ for some closed subgroup $H$? This is a statement in Lang, Real and Functional Analysis, Chpt XII, sec 4. The space below means homogeneous space equipped with $G$ action where $G$ is some topological group.
"Such a space is isomorphic to $G/H$ with some closed subgroup $H$."
$\textbf{Q:}$ Why do I see such $H$ always closed? Say space is $X$. Then $G\to Aut(X)$ is group homomorphism.(Is this even topological group homomorphism?) In particular, why kernel of previous map must be closed? What topological assumption has to be assumed on $Aut(X)$? One can assume $X$ is hausdorff. From $G/H\cong X$, I need $H$ better closed. 

Given a continuous, transitive action $\alpha:G\times X\to X$ of a topological group $G$ on a Hausdorff (or merely $T_1$) space $X$, the $H$ that's used in the statement you quoted is just the stabilizer of any chosen point $x_0\in X$. That is, $H=\{g\in G:\alpha(g,x_0)=x_0\}$. This $H$ is closed because it is the pre-image, under the continuous map $g\to\alpha(g,x_0)$ of the singleton $\{x_0\}$ (which is closed because $X$ is a $T_1$-space). 
(Note that, if we had chosen some other $x_1\in X$ instead of $x_0$, we would have gotten a different closed subgroup $H'$ of $G$, but it would be conjugate to $H$. Indeed, since the action of $G$ on $X$ is transitive, there is some element $t\in G$ with $\alpha(t,x_0)=x_1$, and then $H'=tHt^{-1}$.)
