I'm reading through Atiyah's paper that classifies vector bundles over an elliptic curve, and I'm a little confused about one of his proofs.

Lemma 15(i) states that if $E \in \mathcal{E}(r,d)$ is a vector bundle of rank $r$ and degree $d\geq0$ over $X,$ then $s:=h^0(X,E)=d$ if $d>0$ and $s=0$ or $1$ if $d=0.$

For the proof, if $d\geq r,$ choose a maximal splitting $E=(L_1 ,...,L_r)$ with each $L_i>1$ (where 1 denotes the trivial line bundle). But in Lemma 11 he has the equation

$d=deg(E)=\sum^{r}_{1} deg(L_i).$ If each $L_i>1$ doesn't that mean $deg(L_i )\geq2$ so $deg(E)\geq2r?$ But we are only assuming $d\geq r.$

Anyways, If $deg(L_i)\geq 2 $ then it's clear that $H^1 (X,L_i)=0$. Is that how he claimed that $H^1 (X,L_i)=0$ in the next line or is he using something else?

EDIT: Actually, we don't necessarily have $deg(L_i)\geq 2$ since $1$ denotes the trivial line bundle which has degree $0$. So if $L_i >1$ it just means that $deg(L_i)>0.$ Still, how do we know that $L_i >1$? And how does this imply $H^1 (X,L_i)=0$?

We say that $L_1 \geq L_2$ if $\Gamma Hom(L_2 ,L_1)\neq 0.$ It implies that $deg(L_1)\geq deg(L_2).$

$\Gamma Hom(L_2,L_1) $ is the space of global sections of the bundle $Hom(L_2,L_1)≅L^*_2 \otimes L_1$