Suppose $G$ is a Lie group acting transitively on a manifold $M$. Does that already imply that the connected component $G^\circ$ of $G$ acts transitively on the connected components of $M$? I have a "proof" here but not sure if it is all correct:
Let $x \in M$ and define $G_x$ as the isotropy group of $x$. Since $G$ acts transitively on $M$ we have that $M$ is a homogeneous space and it is diffeomorphic to $G / G_x$. We have that $\pi_x \colon G \to G/G_x=M$ is an open map and since $G^\circ$ is open in $G$, $\pi_x(G^°)$ is open in $M$. But $\pi_x(G^°)$ is the $G^°$-orbit through $x$. Using that for each $x \in M$, we get that $\pi_x(G^°)$ is closed and open and hence $\pi_x(G^°)$ is the union of connected components of $M$. Futhermore $x \in \pi_x(G^°)$, so the component containing $x$ is contained in $\pi_x(G^°)$.
You are correct, but the openness of the action is not obvious. There is the following general result, that can be applied to Lie groups with countably many components:
Theorem. If $G$ is a Hausdorff locally compact group that is a countable union of compact sets acting continuously on a topological space $X$, then for every $x \in X$ the action on $x$ gives a homeomorphism $G/G_x \cong G\cdot x$.
The proof that the map is open uses the Baire Category Theorem for locally compact Hausdorff spaces.
In particular, $G^\circ \cdot x$ is open in $G \cdot x$ for every $x$.