No, this is not true. Consider the tangent bundle to $\mathbb{CP}^2$; I claim it does not contain a sublinebundle. Let $h$ be the class of a hyperplane in $H^2(\mathbb{CP}^2)$. So $H^{\ast}(\mathbb{CP}^2) \cong \mathbb{Z}[h]/h^3$. Here are two proofs (which are really the same proof in two languages).

Proof one: Chern classes The total Chern class of the tangent bundle is $1+3h + 3 h^2$. If we had a short exact sequence $0 \to L_1 \to T \to L_2 \to 0$, then we would have $c_1(L) c_2(L) = 1+3h+3h^2$. But the polynomial $1+3x+3x^2$ is irreducible over $\mathbb{Z}$. $\square$

Proof two: a bundle with no section The projectivation of the tangent bundle is the flag manifold $\mathcal{F}\ell_3$. Choosing a sublinebundle of $T$ corresponds to choosing a (continuous) section of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$. We have $H^{\ast}(\mathcal{F}\ell_3) \cong H^{\ast}(\mathbb{CP}^2)[u]/(u^2+3hu+3h^2)$. If we had a section, this would induce a map from $H^{\ast}(\mathcal{F}\ell_3)$ to $H^{\ast}(\mathbb{CP}^2)$ which would be the identity on $H^{\ast}(\mathbb{CP}^2)$ and would have to send $u$ to $kh$ for some integer $k$. But then we would have $k^2+3k+3=0$, and this polynomial does not have integer roots. $\square$

I think of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$ as a complex analogue of the Hopf fibration, which likewise has no continuous section, although I don't know a way to make this precise. This answer is the analogue of the Hairy Ball theorem, which basically says that the tangent bundle to $S^2$ doesn't have a real sublinebundle.

No, this is not true. Consider the tangent bundle to $\mathbb{CP}^2$; I claim it does not contain a sublinebundle. Let $h$ be the class of a hyperplane in $H^2(\mathbb{CP}^2)$. So $H^{\ast}(\mathbb{CP}^2) \cong \mathbb{Z}[h]/h^3$. Here are two proofs (which are really the same proof in two languages).

Proof one: Chern classes The total Chern class of the tangent bundle is $1+3h + 3 h^2$. If we had a short exact sequence $0 \to L_1 \to T \to L_2 \to 0$, then we would have $c(L_1) c(L_2) = 1+3h+3h^2$. But the polynomial $1+3x+3x^2$ is irreducible over $\mathbb{Z}$. $\square$

Proof two: a bundle with no section The projectivation of the tangent bundle is the flag manifold $\mathcal{F}\ell_3$. Choosing a sublinebundle of $T$ corresponds to choosing a (continuous) section of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$. We have $H^{\ast}(\mathcal{F}\ell_3) \cong H^{\ast}(\mathbb{CP}^2)[u]/(u^2+3hu+3h^2)$. If we had a section, this would induce a map from $H^{\ast}(\mathcal{F}\ell_3)$ to $H^{\ast}(\mathbb{CP}^2)$ which would be the identity on $H^{\ast}(\mathbb{CP}^2)$ and would have to send $u$ to $kh$ for some integer $k$. But then we would have $k^2+3k+3=0$, and this polynomial does not have integer roots. $\square$

I think of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$ as a complex analogue of the Hopf fibration, which likewise has no continuous section, although I don't know a way to make this precise. This answer is the analogue of the Hairy Ball theorem, which basically says that the tangent bundle to $S^2$ doesn't have a real sublinebundle.

No, this is not true. Consider the tangent bundle to $\mathbb{CP}^2$; I claim it does not contain a sublinebundle. Let $h$ be the class of a hyperplane in $H^2(\mathbb{CP}^2)$. So $H^{\ast}(\mathbb{CP}^2) \cong \mathbb{Z}[h]/h^3$. Here are two proofs (which are really the same proof in two languages).

Proof one: Chern classes The total Chern class of the tangent bundle is $1+3h + 3 h^2$. If we had a short exact sequence $0 \to L_1 \to T \to L_2 \to 0$, then we would have $c_1(L) c_2(L) = 1+3h+3h^2$. But the polynomial $1+3x+3x^2$ is irreducible over $\mathbb{Z}$. $\square$

Proof two: a bundle with no section The projectivation of the tangent bundle is the flag manifold $\mathcal{F}\ell_3$. Choosing a sublinebundle of $T$ corresponds to choosing a (continuous) section of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$. We have $H^{\ast}(\mathcal{F}\ell_3) \cong H^{\ast}(\mathbb{CP}^2)[u]/(u^2+3hu+3h^2)$. If we had a section, this would induce a map from $H^{\ast}(\mathcal{F}\ell_3)$ to $H^{\ast}(\mathbb{CP}^2)$ which would be the identity on $H^{\ast}(\mathbb{CP}^2)$ and would have to send $u$ to $kh$ for some integer $k$. But then we would have $k^2+3k+3=0$, and this polynomial does not have integer roots. $\square$

I think of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$ as a complex analogue of the Hopf fibration, which likewise has no continuous section, although I don't know a way to make this precise.

No, this is not true. Consider the tangent bundle to $\mathbb{CP}^2$; I claim it does not contain a sublinebundle. Let $h$ be the class of a hyperplane in $H^2(\mathbb{CP}^2)$. So $H^{\ast}(\mathbb{CP}^2) \cong \mathbb{Z}[h]/h^3$. Here are two proofs (which are really the same proof in two languages).

Proof one: Chern classes The total Chern class of the tangent bundle is $1+3h + 3 h^2$. If we had a short exact sequence $0 \to L_1 \to T \to L_2 \to 0$, then we would have $c_1(L) c_2(L) = 1+3h+3h^2$. But the polynomial $1+3x+3x^2$ is irreducible over $\mathbb{Z}$. $\square$

Proof two: a bundle with no section The projectivation of the tangent bundle is the flag manifold $\mathcal{F}\ell_3$. Choosing a sublinebundle of $T$ corresponds to choosing a (continuous) section of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$. We have $H^{\ast}(\mathcal{F}\ell_3) \cong H^{\ast}(\mathbb{CP}^2)[u]/(u^2+3hu+3h^2)$. If we had a section, this would induce a map from $H^{\ast}(\mathcal{F}\ell_3)$ to $H^{\ast}(\mathbb{CP}^2)$ which would be the identity on $H^{\ast}(\mathbb{CP}^2)$ and would have to send $u$ to $kh$ for some integer $k$. But then we would have $k^2+3k+3=0$, and this polynomial does not have integer roots. $\square$

I think of the bundle $\mathcal{F}\ell_3 \to \mathbb{CP}^2$ as a complex analogue of the Hopf fibration, which likewise has no continuous section, although I don't know a way to make this precise. This answer is the analogue of the Hairy Ball theorem, which basically says that the tangent bundle to $S^2$ doesn't have a real sublinebundle.