You can look at Lieberman's paper Holomorphic Vector Fields on Projective Manifolds.

His proof is more or less as follows. A result of Grothendieck asserts that $\mathrm{Aut}^0(X)$, the connected component of the identity of the automorphism group of $X$, is an algebraic group which acts algebraically on $X$.

Look at the (analytic) subgroup generated by your vector field and let $G$ be its Zariski closure in $\mathrm{Aut}^0(X)$. Notice that $G$ is abelian.

If $p \in X$ is a zero of your vector field then $p$ is fixed by the action of $G$ on $X$. Thus for $k \in \mathbb N$, $G$ acts on $$\frac{\mathcal O_{X,p}}{\mathfrak{m}_p^k}, $$ where $\mathfrak m_p$ is the maximal ideal of $\mathcal O_{X,p}$. Moreover, if $k \gg 0$ then the action is faithfull. Thus $G$ is isomorphic to a linear algebraic group and yet another result of Rosenlicht says that a Zariski-closed abelian subgroup of a linear algebraic group is of the form $(\mathbb Cˆ*, \cdot)^r \times (\mathbb C,+)^s$. The action of the factors of this decomposition generate the sought rational curves.

Added later: For an alternative proof, in a more analytical take, see Theorem 6.4 of this paper. There it is proved that the existence of a non-zero section of $\bigwedge^q TX$ vanishing at a point suffices to ensure that $X$ is uniruled.

You can look at Lieberman's paper Holomorphic Vector Fields on Projective Manifolds.

His proof is more or less as follows. A result of Grothendieck asserts that $\mathrm{Aut}^0(X)$, the connected component of the identity of the automorphism group of $X$, is an algebraic group which acts algebraically on $X$.

Look at the (analytic) subgroup generated by your vector field and let $G$ be its Zariski closure in $\mathrm{Aut}^0(X)$. Notice that $G$ is abelian.

If $p \in X$ is a zero of your vector field then $p$ is fixed by the action of $G$ on $X$. Thus for $k \in \mathbb N$, $G$ acts on $$\frac{\mathcal O_{X,p}}{\mathfrak{m}_p^k}, $$ where $\mathfrak m_p$ is the maximal ideal of $\mathcal O_{X,p}$. Moreover, if $k \gg 0$ then the action is faithfull. Thus $G$ is isomorphic to a linear algebraic group and yet another result of Rosenlicht says that a Zariski-closed abelian subgroup of a linear algebraic group is of the form $(\mathbb Cˆ*, \cdot)^r \times (\mathbb C,+)^s$. The action of the factors of this decomposition generate the sought rational curves.

Added later: For an alternative proof see Theorem 6.4 of this paper. There it is proved that the existence of a non-zero section of $\bigwedge^q TX$ vanishing at a point suffices to ensure that $X$ is uniruled.

You can look at Lieberman's paper Holomorphic Vector Fields on Projective Manifolds.

His proof is more or less as follows. A result of Grothendieck asserts that $\mathrm{Aut}^0(X)$, the connected component of the identity of the automorphism group of $X$, is an algebraic group which acts algebraically on $X$.

Look at the (analytic) subgroup generated by your vector field and let $G$ be its Zariski closure in $\mathrm{Aut}^0(X)$. Notice that $G$ is abelian.

If $p \in X$ is a zero of your vector field then $p$ is fixed by the action of $G$ on $X$. Thus for $k \in \mathbb N$, $G$ acts on $$\frac{\mathcal O_{X,p}}{\mathfrak{m}_p^k}, $$ where $\mathfrak m_p$ is the maximal ideal of $\mathcal O_{X,p}$. Moreover, if $k \gg 0$ then the action is faithfull. Thus $G$ is isomorphic to a linear algebraic group and yet another result of Rosenlicht says that a Zariski-closed abelian subgroup of a linear algebraic group is of the form $(\mathbb Cˆ*, \cdot)^r \times (\mathbb C,+)^s$. The action of the factors of this decomposition generate the sought rational curves.

You can look at Lieberman's paper Holomorphic Vector Fields on Projective Manifolds.

His proof is more or less as follows. A result of Grothendieck asserts that $\mathrm{Aut}^0(X)$, the connected component of the identity of the automorphism group of $X$, is an algebraic group which acts algebraically on $X$.

Look at the (analytic) subgroup generated by your vector field and let $G$ be its Zariski closure in $\mathrm{Aut}^0(X)$. Notice that $G$ is abelian.

If $p \in X$ is a zero of your vector field then $p$ is fixed by the action of $G$ on $X$. Thus for $k \in \mathbb N$, $G$ acts on $$\frac{\mathcal O_{X,p}}{\mathfrak{m}_p^k}, $$ where $\mathfrak m_p$ is the maximal ideal of $\mathcal O_{X,p}$. Moreover, if $k \gg 0$ then the action is faithfull. Thus $G$ is isomorphic to a linear algebraic group and yet another result of Rosenlicht says that a Zariski-closed abelian subgroup of a linear algebraic group is of the form $(\mathbb Cˆ*, \cdot)^r \times (\mathbb C,+)^s$. The action of the factors of this decomposition generate the sought rational curves.

Added later: For an alternative proof, in a more analytical take, see Theorem 6.4 of this paper. There it is proved that the existence of a non-zero section of $\bigwedge^q TX$ vanishing at a point suffices to ensure that $X$ is uniruled.