I proved the following result
"Prove that if $V$ is a differentiable vector field on a compact surface $S$ and $\alpha(t)$ is the maximal trajectory of $V$ with $\alpha(0)=p\in S$, then $\alpha(t)$ is defined for all $t\in\mathbb{R}.$"
For every $q\in S$, there exist a neighborhood $B$ of $q$ and an interval $(-\varepsilon,\varepsilon)$ such that the trayectory $\alpha(t)$, with $\alpha(0)=q$, is defined in $(-\varepsilon,\varepsilon)$. By compactness, it's possible to cover $S$ with a finite numbers of such neighborhoods, ${B_j}'s$. Let $\varepsilon_0=\min\{\varepsilon_j\}$. If $\alpha(t)$ is defined for $t<t_0$ and isn't defined for $t_0$, we take $t_1\in (0,t_0)$, with $|t_0-t_1|<\varepsilon/3$. Consider the trayectory $\gamma(t)$ of $V$, with $\gamma(0)=\alpha(t_1)$, which leads us to a contradiction.
My question: Is the result valid if we have complete surface instead of compact surface? And if this is true (which I believe). How could it be demonstrated? I have tried but have not been successful.
As Arctic Char pointed out, your quetion has nothing to do with any Riemannian structure. Here is an example of a complete Riemannian manifold $(M,g)$ which has a vector field with an integral curve that is not defined for all time.
Let $(\mathbb{R}^2,g)$ be the usual Euclidean plane. It is complete.
Let $f\colon \mathbb{R}^2 \to \mathbb{S}^2 \setminus \{N\}$ be the standard diffeomorphism given by the sterographic projection from the north pole of the two sphere.
Let $X$ be any vector field on the $2$ sphere that is locally non-zero around the north pole $N$. Then there exists a non-constant integral curve of $X$ passing through $N$, say $\sigma$, at a time $t_0$.
Then $V = f^*X$ is a vector field of $(\mathbb{R}^2,g)$ and $f^*\sigma$ is an integral curve of $V$ which is not defined for $t = t_0$.
Comment Compactness is a topological property, while completeness is a metric property. Of course, every metric on a compact manifold is complete, but compactness results are not induced by metric statements: one couldn't expect your thought to be true because of that.