To show $X$ is a complete vector field on $M$ Well, I have solved myself the problem : every smooth vector field on a compact manifold is complete.
Now I have got this problem which I am not able to progress:
let $X$ is a vector field on $M$, suppose $\exists \epsilon >0\ni (-\epsilon,\epsilon) \subsetneq (a(m),b(m))\forall m\in M$, Then show that $X$ is a complete vector field i.e $(a(m), b(m))=\mathbb{R}$
thank you for any help.
where $(a(m),b(m))$ is the domain of maximal integral curve, for each $m\in M$ we will get this interval $(a(m),b(m))$ 
here I asked some realted things 
A series of Lemmas about $C^{\infty}$ vector fields
 A: Let $\gamma(t)$ be the maximal integral curve with initial condition $\gamma(0) = m$, and let's suppose $b(m) < \infty$.
If we denote $\gamma'(t)$ the maximal integral curve with initial condition $\gamma'(0) = \gamma(b(m) -\epsilon/2)$ then
$$
\gamma''(t) =
\begin{cases}
\gamma(t) & \text{if } t\in (a(m), b(m) - \epsilon/2)\\
\gamma'(t - b(m) + \epsilon/2)& \text{if } t\in [b(m) - \epsilon/2, b(m) + \epsilon/2)
\end{cases}
$$
prolongs $\gamma(t)$. That contradicts the assertion that $\gamma(t)$ is maximal.
A similar reasoning can be used to show $a(m) = -\infty$.
Edit - Some clarifications
$\gamma''$ prolongs $\gamma$ means that $\gamma''$ has an interval of existence, $(a(m), b(m) + \epsilon/2)$, that is a proper superset of the interval of existence, $(a(m), b(m))$, of $\gamma$ and the two curves are equal on the common domain, $(a(m), b(m))$.
The "uniformity" of $\epsilon$ ensures that the integral curve with initial condition 
$$
\gamma'(0) = \gamma(b(m) - \epsilon/2)
$$
exists on the interval $[0, \epsilon)$. Without "uniformity" it could happen that the maximal right interval of existence of $\gamma'$ is $[0, \epsilon/2)$. In such a case, a curve $\gamma''$, constructed as above, would coincide with $\gamma$: it would not prolong $\gamma$.
The initial condition of $\gamma'$ is chosen in such a way to allow us to smoothly join $\gamma'$ to $\gamma$ in order to form a new curve, $\gamma''$, that prolongs $\gamma$.
The contradiction arises because we constructed an integral curve that prolongs a maximal integral curve, which, as such, cannot be prolonged by definition. 
A: For any $m\in M$, write $\mathrm{Fl}^X_t(m)$ for the maximal integral curve starting at $m$ at $t=0$, so $\mathrm{Fl}^X(m):(a(m),b(m))\rightarrow M,~\mathrm{Fl}^X_0(m)=m$ and $\frac{d}{dt}\mathrm{Fl}^X_t(m)=X_{\mathrm{Fl}^X_t(m)}$.
We use the fact that
$$\Omega=\lbrace (m,t)\in M\times \mathbb R~|~\mathrm{Fl}^X_t(m)~\mathrm{exists}~\rbrace$$
is an open subset of $M\times \mathbb R$ containing $M\times 0$. Consider an integral curve (not necessarily maximal) of $X$ that starts at $m$ and is defined on $(a,b)$ with $-\infty\leq a<0<b<+\infty$. By compactness of $M$, there is a sequence $t_k\rightarrow b$ with $$\mathrm{Fl}^X_{t_k}(m)\rightarrow m'.$$
By what was said about $\Omega$, there exists $\epsilon>0$ and an open neighborhood $U$ of $m'$ with $U\times (-\epsilon,+\epsilon)\subset \Omega.$ Choose $k$ large enough so that $b-\epsilon/2< t_k <b$ and $\mathrm{Fl}^X_{t_k}(m)\in U$. We set $m_k=\mathrm{Fl}^X_{t_k}(m)$. By construction of $U$ and $\epsilon$, the integral curve starting at $m_k$ is defined on an interval containing $(-\epsilon,+\epsilon)$. Now we define the following curve:
$$\gamma:(a,t_k+\epsilon)\rightarrow M,t\mapsto\bigg(
\begin{array}{ll}
\underline{t\in (a,t_k]:} & \mathrm{Fl}^X_t(m) \\
\underline{t\in [t_k,t_k+\epsilon):} & \mathrm{Fl}^X_{t-t_k}(m_k)
\end{array}$$
This curve is smooth, and prolongs the integral curve to $(a,t_k+\epsilon)\supset (a,b+\epsilon/2)$. To check smoothness, you only need to look at the curve around $t_k$. It then follows from the uniqueness of integral curves that $\mathrm{Fl}^X_t(m)$ and $\mathrm{Fl}^X_{t-t_k}(m_k)$ coincide on a neighborhood of $t_k$, which shows that $\gamma$ is smooth at $t_k$.
Thus, the maximal integral curves are defined for all times in both directions (upon switching to $-X$.)
