proof that on a compact manifold a vector field is complete I would like to know if my proof of the following statement is correct

If $M$ is a compact manifold, then every vector field $X$ over $M$ is complete.

Proof:
I take $p\in M$ and $(\gamma_p,I_p)$ a maximal integral curve through $p$. $I_p$ is open and non empty so if $I_p$ is closed, it is equal to $\mathbb{R}$.
So take $s\in \text{Adh}(I_p)$. There exists $t_i\rightarrow s, t_i\in I_p$. Define $q$ as $\text{lim}(\gamma_p(t_i))$ and take a maximal integral curve passing by $q$ as $(\gamma_q,I_q)$. Then since the vector fields are $C^\infty$, $\gamma_q(I_q)\cap\gamma_p(I_p)$ is nonempty (intuitively, this is clear but don't really know how to show it).
On the intersection, $\gamma_p$ and $\gamma_q$ coincide. So we can extend the maximal curve $\gamma_p$ to include $\gamma_q$. This implies that $q\in\gamma_p(I_p)$ and is actually equal to $\gamma_p(s)$
I think the main idea is there even though it probably lacks a bit of rigour.
 A: As Jack Lee is already addressing the flaws of your proof, I thought I would drop a complete proof of the statement, so it is at hand for everyone having a question on it.
First, let us work out a general statement:

Proposition. Let $M$ be a smooth manifold and $X$ be a vector field on $M$ with a local flow given by $(\phi_t)_t$. Assume that there exists $\varepsilon>0$ such that $\phi$ is defined on $]-2\varepsilon,2\varepsilon[\times M$, then $X$ is complete.

Proof. For all $t\in\mathbb{R}$, let $k(t)$ be the integer part of $t/\varepsilon$, then one has:
$$t-k(t)\varepsilon\in[-\varepsilon,0]\subseteq]-2\varepsilon,2\varepsilon[,$$
so that one can define the following diffeomorphism of $M$:
$$\psi_t:={\phi_{\varepsilon}}^{k(t)}\circ\phi_{t-k(t)\varepsilon}.$$
For all $x\in M$, since $k(0)=0$, one has:
$$\psi_0(x)=\phi_0(x)=x.$$
Furthermore, for all $s\in\mathbb{R}$, one has the following equality:
$$\begin{align}\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert t=s}\psi_t(x)&=\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert t=s}{\phi_{\varepsilon}}^{k(s)}\circ\phi_{t-k(s)\varepsilon}(x),\\&=\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert t=s-k(s)\varepsilon}{\phi_{\varepsilon}}^{k(s)}\circ\phi_t(x),\\&=T_{\phi_{s-k(s)\varepsilon}(x)}{\phi_{\varepsilon}}^{k(s)}\left(\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert t=s-k(s)\varepsilon}\phi_t(x)\right),\\&=T_{\phi_{s-k(s)\varepsilon}(x)}{\phi_{\varepsilon}}^{k(s)}(X(\phi_{s-k(s)\varepsilon}(x))),\\&=T_{\phi_{\varepsilon}(\phi_{s-k(s)\varepsilon}(x))}{\phi_{\varepsilon}}^{k(s)-1}(T_{\phi_{s-k(s)\varepsilon}(x)}\phi_{\varepsilon}(X(\phi_{s-k(s)\varepsilon}(x)))),\\&=T_{\phi_{\varepsilon}(\phi_{s-k(s)\varepsilon}(x))}{\phi_{\varepsilon}}^{k(s)-1}(X(\phi_{\varepsilon}(\phi_{s-k(s)\varepsilon}(x)))),\\&=T_{{\phi_{\varepsilon}}^{k(s)}(\phi_{s-k(s)\varepsilon}(x))}{\phi_{\varepsilon}}^0(X({\phi_{\varepsilon}}^{k(s)}(\phi_{s-k(s)\varepsilon}(x)))),&\\&=X(\psi_s(x)).\end{align}$$
Therefore, by the unicity part of Picard-Lindelöf theorem, $\phi=\psi$ and $\phi$ is in fact defined on $\mathbb{R}\times M$. Whence the result. $\Box$
Remark. The key point of these computations is that $\phi$ preserves $X$, for all $t$ such that $\phi_t$ exists and $x\in M$:
$$T_x\phi_t(X(x))=X(\phi_t(x)).$$
Which is almost tautological, since by the very definition of the flow, one has:
$$X(\phi_t(x))=\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert s=t}\phi_s(x)=\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert s=0}\phi_t\circ\phi_s(x)=T_{\phi_0(x)}\phi_t\left(\frac{\mathrm{d}}{\mathrm{d}t}_{\big\vert s=0}\phi_s(x)\right)=T_x\phi_t(X(x)).$$
Another thing to notice is that for all $t$ sufficiently close to $s$, $k(t)=k(s)$.
From there it is easy to derive the desired result.

Corollary. Let $M$ be a compact smooth manifold and $X$ be a vector field on $M$, then $X$ is complete.

Proof. Let $p\in M$, using the existence part of Picard-Lindelöf theorem, there exists $\varepsilon_p>0$ and $U_p$ an open neighborhood of $p$ in $M$ such that $\phi$ the flow of $X$ is defined on $]-\varepsilon_p,\varepsilon_p[\times U_p$. By construction, $\{U_p\}_{p\in M}$ is an open cover of $M$, which is compact, hence there exists $p_1,\ldots,p_k$ in $M$ such that $\{U_{p_i}\}_{1\leqslant i\leqslant k}$ is still a cover of $M$. Let then define the following existence time:
$$\varepsilon:=\min_{1\leqslant i\leqslant k}\varepsilon_{p_i}>0,$$
by construction, for all $i\in\{1,\ldots,k\}$, $\phi$ is defined on $]-\varepsilon,\varepsilon[\times U_i$, therefore on the whole $]-\varepsilon,\varepsilon[\times M$. Whence the result using the above proposition. $\Box$
