Second countability is invariant under orbit space of an action I need to prove that if $X$ is a second countable space and $f:G\times X\to X$ is a left action on $X$, then its orbit space $X/G$ is also second countable. Here is my idea:
Let $\mathscr{B}$ be a countable basis for $X$, for each $B\in \mathscr{B}$ let $f(G,B)$ denote the set of all orbits $$[x]=\{z\in X\mid z=f(g,x) \text{ for all }g\in G \}$$ with $x\in B$. I want to show that the collection of all sets $f(G,B)$ for all $B\in \mathscr{B}$ is a basis for $X/G$. How could I go about it?
 A: (I think you meant to write $[x]=\{z\in X|z=f(g,x)$ for some $g\in G\}$, otherwise this isn't really the orbit of $x$).
You idea seems fine to me. You can prove this is a basis as follows. First, one needs to show that its elements are actually open sets. This follows from the following lemma.
Lemma. If a group $G$ acts on a topologixal space $X$, then the quotient map $X\to X/G$ is open.
Proof. The proof goes roughly as follows. Suppose $U\subseteq X$ is open. We want to show $p(U)$ is open in $X/G$. Therefore by definition of the quotient topology, we need to show $p^{-1}(p(U))$ is open. But $p^{-1}(p(U))=G.U=\{gu|g\in G, u\in U\}= \bigcup_{g\in G}gU$, which is a union of open sets, and hence open.
Now, we need to show every open set in $X/G$ is a union of sets from $\{f(G,B)|B\in\mathscr{V}\}$. Denote by $p:X\to X/G$ the natural quotient map. Let $U\subseteq X/G$ be an open set. By definition of the quotient topology, this means its preimage in X, i.e. $p^{-1}(U)$, is open in $X$. Therefore $p^{-1}(U)=\bigcup_{B\in \mathcal{B}}B$ for some subset $\mathcal{B}\subseteq \mathscr{B}$. Then it's easy to show $U=\bigcup_{B\in \mathcal{B}} f(G,B)$, as needed.
In general, an image of a second-countable space under a continuous and open map is second countable. And it is always the case that the quotient map under an action of a group is continuous and open.
