Compactly supported smooth vector fields are complete

Let $\gamma :I\to M$ be a maximal integral curve of $X$ through some point $p$. If $\gamma (t_0)\mathrm{\notin }K$ for some $t_0\in I$, then $X_{\gamma (t_0)}=0$ by [L3], so the constant curve through $\gamma (t_0)$ is an integral curve of $X$. Uniqueness therefore forces $\gamma$ to be constant on the connected component of $\{t\in I:\gamma (t)\mathrm{\notin }K\}$ containing $t_0$. Thus every nonconstant part of $\gamma$ lies in $K$.

Suppose $I$ had a finite right endpoint $b$. Choose times $t_n\uparrow b$. If infinitely many $\gamma (t_n)$ lie in $K$, compactness of $K$ gives a subsequence converging to some $q\in K$. Otherwise $\gamma (t_n)\mathrm{\notin }K$ for all large $n$, and step 1.1 makes those tail values constant on a neighbourhood of $b$; hence $\gamma (t_n)\to q$ for some $q\in M\setminus K$.

By [L2], there is a local flow through $q$ defined on some interval $(-\epsilon ,\epsilon )$. For $n$ large, $\gamma (t_n)$ lies in its domain, so uniqueness of integral curves extends $\gamma$ past $b$ by flowing forward from $\gamma (t_n)$ for time larger than $b-t_n$. This contradicts maximality.

The same argument excludes a finite left endpoint. Therefore every maximal integral curve is defined on all of $\mathbb{R}$, and [L1] implies that $X$ is complete.