Prove that if $M$ is a connected, smooth manifold, then $\text{Diff}(M)$ acts transitively on $M$.
What I'm trying to do is to take an arbitrary $p\in M$ and show that the orbit of $p$
$$\mathcal{O}_p:=\{q\in M \mid \phi(q)=p \text{ for some }\phi\in\text{Diff}(M)\}$$
is both open and closed. Since $M$ is connected, $\mathcal{O}_p=M$ for any $p$, which proves the proposition.
But I'm having difficulty to prove $\mathcal{O}_p$ is open in the first place.
My initial idea was to take a chart $(U,\psi)$ about $p$ and, for any $q\in U$, take some $g\in\text{Diff}(\mathbb{R}^n)$ such that $g(\psi(q))=\psi(p)$ (that's possible because $\text{Diff}(\mathbb{R}^n)$ is transitive). Then
$$\phi:=\psi^{-1}\circ g\circ \psi \in \text{Diff}(U)$$
has the property that $\phi(q)=p$. But that is not enough, since we need $\phi\in\text{Diff}(M)$. I've tried to prove that there is an extension of $\phi\in\text{Diff}(U)$ to some $\bar{\phi}\in\text{Diff}(M)$, but I couldn't.
Besides, I have no idea how to prove $\mathcal{O}_p$ is closed.
Any ideas? Thanks!