[proofplan] Fix a basepoint $p \in M$ and let $U$ denote the orbit of $p$ under $\mathrm{Diff}(M)$. We show that $U$ is both open and closed in $M$; since $M$ is connected and $U$ is non-empty, this forces $U = M$. The key local ingredient is that in a Euclidean chart, any point in the unit ball can be moved to the origin by a compactly supported diffeomorphism obtained by flowing along a bump-cut-off constant vector field. [/proofplan]

[step:Set up the orbit $U$ of $p$ and reduce to showing it is open and closed] Fix $p \in M$ and define \begin{align*} U &:= \{\, q \in M : \exists\, \psi \in \mathrm{Diff}(M) \text{ with } \psi(p) = q \,\}. \end{align*} Then $p \in U$ since $\mathrm{id}_M \in \mathrm{Diff}(M)$, so $U \neq \varnothing$. Because $M$ is connected, to conclude $U = M$ it suffices to prove that $U$ is open and that $M \setminus U$ is open. [/step]

[step:Construct a compactly supported diffeomorphism of $\mathbb{R}^n$ moving $0$ to any target in $B(0,1)$] We prove the following local ingredient. For every $a \in B(0,1) \subset \mathbb{R}^n$ there exists $\psi_a \in \mathrm{Diff}(\mathbb{R}^n)$ with $\psi_a(0) = a$ and $\mathrm{supp}(\psi_a) \subset \overline{B}(0,2)$, where $\mathrm{supp}(\psi_a) := \overline{\{x \in \mathbb{R}^n : \psi_a(x) \ne x\}}$. Choose a smooth cut-off $\chi: \mathbb{R}^n \to \mathbb{R}$ with $\chi \in C_c^\infty(\mathbb{R}^n)$, $0 \le \chi \le 1$, $\chi \equiv 1$ on $\overline{B}(0,1)$, and $\chi \equiv 0$ outside $B(0,2)$. Define the vector field \begin{align*} V_w: \mathbb{R}^n &\to \mathbb{R}^n \\ x &\mapsto \chi(x)\, w \end{align*} for a fixed direction $w \in \mathbb{R}^n$. Because $V_w$ is smooth with compact support, it is complete, and its flow $\varphi^{V_w}_t: \mathbb{R}^n \to \mathbb{R}^n$ is defined for all $t \in \mathbb{R}$ and is a diffeomorphism of $\mathbb{R}^n$ for each $t$. For $|x| \ge 2$ the field vanishes, so $\varphi^{V_w}_t(x) = x$; hence $\mathrm{supp}(\varphi^{V_w}_t) \subset \overline{B}(0,2)$. Given $a \in B(0,1)$, set $w := a / \|a\|$ if $a \ne 0$ and let $\psi_a := \varphi^{V_w}_{\|a\|}$; if $a = 0$, take $\psi_a := \mathrm{id}$. Since $\chi \equiv 1$ on $\overline{B}(0,1)$, on the line segment $\{tw : 0 \le t \le \|a\|\} \subset \overline{B}(0,1)$ the field $V_w$ restricts to the constant $w$, so the integral curve of $V_w$ through the origin is $t \mapsto tw$ for $t \in [0, \|a\|]$. Therefore $\psi_a(0) = \varphi^{V_w}_{\|a\|}(0) = \|a\|\, w = a$. [/step]

[step:Transfer the local ingredient to any chart via pushforward] Let $r \in M$ and let $(W, \varphi)$ be a smooth chart about $r$ with $\varphi(r) = 0$, $\varphi(W) \supset \overline{B}(0,2)$. Let $W_0 := \varphi^{-1}(B(0,1)) \subset W$, an open neighbourhood of $r$ in $M$. For any $r' \in W_0$, write $a := \varphi(r') \in B(0,1)$. By the previous step, there exists $\psi_a \in \mathrm{Diff}(\mathbb{R}^n)$ with $\psi_a(0) = a$ and $\mathrm{supp}(\psi_a) \subset \overline{B}(0,2) \subset \varphi(W)$. Define \begin{align*} \Psi_{r,r'}: M &\to M \\ x &\mapsto \begin{cases} \varphi^{-1}\!\bigl(\psi_a(\varphi(x))\bigr) & x \in W, \\ x & x \in M \setminus \varphi^{-1}(\overline{B}(0,2)). \end{cases} \end{align*} The two definitions agree on the overlap $W \setminus \varphi^{-1}(\overline{B}(0,2))$ because $\psi_a$ is the identity outside $\overline{B}(0,2)$, and the set $\varphi^{-1}(\overline{B}(0,2))$ is compact in $W$ (as the continuous image of a compact set under $\varphi^{-1}$), hence closed in $M$. Therefore $\Psi_{r,r'}$ is well-defined and smooth on each of the two open pieces whose union is $M$, and is a diffeomorphism: its inverse is given by the analogous construction using $\psi_a^{-1}$. By construction, $\Psi_{r,r'}(r) = r'$. [/step]

[step:Deduce that $U$ is open using the local ingredient at points of $U$] Let $r \in U$ and pick $\psi \in \mathrm{Diff}(M)$ with $\psi(p) = r$. Choose the neighbourhood $W_0 \ni r$ from the previous step. For any $r' \in W_0$, form $\Psi_{r,r'} \in \mathrm{Diff}(M)$ with $\Psi_{r,r'}(r) = r'$. Then \begin{align*} (\Psi_{r,r'} \circ \psi)(p) &= \Psi_{r,r'}(r) = r', \end{align*} and $\Psi_{r,r'} \circ \psi \in \mathrm{Diff}(M)$. Hence $r' \in U$, so $W_0 \subset U$. Since every $r \in U$ has such a neighbourhood, $U$ is open. [/step]

[step:Deduce that $M \setminus U$ is open using the local ingredient at points outside $U$] Let $r \in M \setminus U$ and choose the neighbourhood $W_0 \ni r$. Suppose for contradiction that $W_0 \not\subset M \setminus U$; then there exists $r' \in W_0 \cap U$. By the previous step applied with roles swapped, there is $\Phi \in \mathrm{Diff}(M)$ with $\Phi(r') = r$ (take $\Psi_{r',r}$ obtained by choosing a chart centred at $r'$, or equivalently invert the map $\Psi_{r,r'}$ constructed above, which sends $r \mapsto r'$, to obtain a diffeomorphism sending $r' \mapsto r$). Since $r' \in U$, there exists $\psi \in \mathrm{Diff}(M)$ with $\psi(p) = r'$. Then \begin{align*} (\Phi \circ \psi)(p) &= \Phi(r') = r, \end{align*} and $\Phi \circ \psi \in \mathrm{Diff}(M)$, forcing $r \in U$, a contradiction. Therefore $W_0 \subset M \setminus U$, and $M \setminus U$ is open. [/step]

[step:Conclude transitivity by connectedness of $M$] The set $U$ is non-empty, open, and has open complement; hence $U$ is clopen in $M$. Connectedness of $M$ forces $U = M$. Therefore, for every $q \in M$ there exists $\psi \in \mathrm{Diff}(M)$ with $\psi(p) = q$. Since the basepoint $p$ was arbitrary and composition of diffeomorphisms is a diffeomorphism, for any pair $p, q \in M$ we obtain $\psi, \psi' \in \mathrm{Diff}(M)$ with $\psi(p_0) = p$, $\psi'(p_0) = q$ for a fixed $p_0 \in M$, and $\psi' \circ \psi^{-1}$ sends $p$ to $q$. This proves that $\mathrm{Diff}(M)$ acts transitively on $M$. [/step]