This is an exercise in Newstead's notes "Vector bundles on algebraic curves".
Let $E$ be a semistable bundle over a genus $g$ complex curve, i.e. for any proper subbundle $F \subset E$ we have $\mathrm{deg} F / \mathrm{rank} F \le \mathrm{deg} E / \mathrm{rank} E$. Let $r$ and $d$ denote the rank and degree of $E$, respectively.
Claim: If $d > r(2g-1)$, then $E$ is globally generated. Moreover, $H^1(E)=0$.
So far I know the following: $H^1(E)=0$ by application of Serre duality and computing the degree of $E^* \otimes K$ for the canonical bundle $K$. To show global generation, I have thought about using the semistability condition for the kernel subbundle $F$ of the map from $E$ to the cokernel coherent sheaf $C$ of the evaluation map $\mathcal{O}^{h^0(E)} \to E$, or computing the number of global sections for $E(-p)$ for a closed point $p$ on the curve. But somehow I am stuck with either approach.
Using the fact that $E$ is semistable, we can deduce that $E(-p)$ is also semistable, as semistability is not influenced by tensoring with a line bundle. Hence, $H^1(E(-p)) = 0$ (by Serre duality and $\nexists E \to K$ if $\deg E > \deg K$). The following sequence is exact: $$ 0 \to E(-p)\to E \to E|_p\to 0 $$ So $$ H^0(E) \to H^0(E|_p) \cong E_p\otimes k(x) $$ is surjective. By Nakayama's Lemma we have $H^0(E)\otimes\mathcal{O} \to E$ is surjective.