I would like to prove that if $H\subset G$ is a normal amenable subgroup such that $G/H$ is amenable, then $G$ is amenable. The definition of amenability that I'm using is the following:
A group $G$ is amenable if every action of $G$ by homeomorphisms of a compact metric space admits an invariant probability measure.
This definition can be found on Navas's "Groups of Circle Diffeomorphisms". I've tried a lot of different ways but i couldn't prove it, I know there are many equivallent definitions for amenability but I would like (if possible) a proof that only uses this definition.
Here is what I've done so far: If $G$ acts on $(M,d)$ then $G/H$ acts on $M/H$ (the quotient of $M$ by the orbits of $H$), the problem is that this group is not necessarily metric, whe could endow the quotient group with the pseudometric $d'$ given in wikipedia https://en.wikipedia.org/wiki/Metric_space#Quotient_metric_spaces (the topology could be weaker than the quotient topology), and then do another quotient $X=(M/H)/\sim$ where $[x]\sim [y]$ if $d'([x],[y])=0$. Here $X$ is a compact metric space and we could take the action of $G/H$ on $X$ given by ${[g]}({[[x]]})=[[y]]$ if $[[g(x)]]=[[y]]$, since $G/H$ is amenable there exists an invariant probability measure, namely $\nu$. Now the sets $A_{[[x]]}=\lbrace y\in M:[[y]]=[[x]]\rbrace$ are compact and invariant under the action of $H$, so each one has an invariant probability measure namely $\mu_{[[x]]}$ and we could define the probability measure on $M$ as $$\mu(B)=\int_X \mu_{[[x]]}(B\cap A_{[[x]]})d\nu([[x]]).$$
I don't know if this works in general, i couldn't prove or disprove it, i suppose this doesn't work since there could be some internal shifting of the orbits of $H$ in the sets $A_{[[x]]}$, but I hope this gives you some insight of what I'm trying so far.
I hope I was clear, many thanks in advance.
Something that might help: The space of probability measures on a metric space is compact, so you could use convergence of probability neasures.
Fix a compact metric space $M.$ Let $W(M)$ denote the Wasserstein space for $M$: the space of probability measures on $M,$ with the Wasserstein metric. The important property is that this metric gives the topology of weak convergence, making $W(M)$ a compact metric space.
Let $W(M)^H$ denote the subspace of $H$-invariant measures. This is closed, so it's also a compact metric space.
An action of $G$ on $M$ gives an action $(gp)(A)=p(g^{-1}A)$ on $W(M).$ Since $H$ is normal, $G$ preserves $W(M)^H$: if $p$ is $H$ invariant then $p(g^{-1}hA)=p((g^{-1}hg)g^{-1}A)=p(g^{-1}A).$ But $H$ acts trivially on $W(M)^H,$ so in fact $G/H$ acts on $W(M)^H.$ Since $G/H$ is amenable there's a $G$-invariant measure $\xi$ on $W(M)^H.$
This is a probability measure on a space of probability measures. To get a measure on the original space $M,$ we need integration of measures. Or in other words the multiplication of the Kantorovich monad. Define $E\xi\in W(M)$ by $(E\xi)(A)=\int p(A)d\xi(p)$ for each Borel $A.$ The $G$-invariance of $\xi$ implies the $G$-invariance of $E\xi$: $$(gE\xi)(A)=\int (gp)(A)d\xi(p)=(E\xi)(A).$$
Finally I would like to mention that the same argument works if you drop the metrizability condition everywhere. The existence of an invariant probability measure for every $G$-action on a compact Hausdorff space is one of the few definitions of amenability that generalizes usefully to non-locally compact groups.