Hartshorne Exercise IV.3.7.
If $X$ is a curve of genus 2, show that a divisor $D$ is very ample $\iff$ $\deg D\ge 5$.
Let $D$ be a divisor on a smooth projective curve $X$. Recall that $D$ is very ample iff $h^0(\OO_X(D-p-q))=h^0(\OO_X(D))-2$ for all $p,q\in X$. One direction is immediate from the fact that if $\deg D\ge 2g+1$ then $D$ is very ample. For the other direction, assume that $D$ is very ample. $X$ has genus 2, and $D$ gives an embedding into projective space of dimension $N=h^0(\OO_X(D)-1$, therefore $N$ must be at least $3$ because $\PP{}{1}$ has genus $0$ and $\PP{k}{2}$ contains no curves of genus 2 (indeed, $\frac{(d-1)(d-2)}{2}\in \{0,1\}\cup[3,\infty)$). So, $h^0(\OO_X(D))\ge 4$. By our first remark, this implies that for any $p,q\in X$, $h^0(\OO_X(D-p-q))\ge2$; in particular, $\deg D\ge 2=\deg K_X$. By RR,
$$\deg D=h^0(\OO_X(D))-h^1(\OO_X(D))+1.$$
Since $\deg D\ge \deg K_X$, either $h^1(\OO_X(D))=0$ or $\deg D=\deg K_X$ and in this case, if $h^0(\OO_X(K_X-D)\ne 0$ then $D\sim K_X$ which contradictions that $h^0(\OO_X(D))\ge 4$ because $H^0(X,\omega_X)\cong k^{g(X)}$. Thus, we see in all cases that $H^1(X,\OO_X(D))=0$, and we conclude that $\deg D =h^0(\OO_X(D))+1\ge 5$, as desired.
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