Solution to DE over closed smooth manifold I am trying to prove the following assertion:
Let $V$ be a smooth vector field on a closed manifold. Then the differential equation $x'=V(x)$ has a solution for all time.
My idea is to use local existence around each point, and by compactness pick a subcover of the open cover formed by these neighborhoods and "patch" them together. Is this correct?
 A: For each $x \in M$, choose $\varphi_x:U_x \to M$ a chart such that $x \in \varphi_x(U_x)$.  Let $x \in W_x \subset K_x \subset U_x$ where $K_x$ is compact, and $V_x$ is open.  Choose a finite subcover $\varphi_{x_1}(W_{x_1}),\dots,\varphi_{x_n}(W_{x_n})$ of $M$.  Let $\tilde V_k$ be the vector field on $U_{x_k}$ such that its push forward by $\varphi_{x_k}$ is $V$ on $\varphi_{x_k}(U_{x_k})$.  Let $\delta>0$ be such that for all $y \in M$, there exists $k = 1,\dots,n$ with $B(\varphi_{x_k}^{-1}(y),\delta) \subset V_{x_k}$.  Let $\Delta = \max_k \sup\{|V_{x_k}(z)|: z \in K_{x_k}\}$.
Let $T$ be the supremum of those $\tau$ such that there exists a solution $x:[0,\tau] \to M$ satisfying $\dot x = V(x)$.  Suppose $T < \infty$, for a contradiction.
Choose $\tau > \max\{0,T - \frac12\delta/\Delta\}$.  Solve the equation $\dot z = \tilde V_k(z)$, $z(\tau) = \varphi_{x_k}^{-1} (x(\tau))$ on $W_{x_k}$.  Note that it has a solution for $t \in [\tau,T+\frac12\delta/\Delta]$.  Push forward this to a solution $\dot x = V(x)$ on $M$ for $t \in [0,T+\epsilon]$.
A: The theorem about the maximal solution tells us that if one end of its domain interval is finite, then the solution will leave every compact subset of the domain of the ODE when approaching this interval end.
As the manifold as domain of the ODE is itself compact, no solution can leave this compact set, thus by contradiction the domain interval of the maximal solution has no finite end-point, it is the whole real line.
