There is serious flaw in the following argument, but I have yet to see what it is:
Let $C$ be a genus $g$ hyperelliptic curve in an abelian variety $A$ (over an algebraically closed field with characteristic $\neq$2), such that the automorphism $-1$ of $A$ descends to the hyperelliptic involution on $C$. The 2:1 projection $C\to \mathbb{P}^1$ is just the image of $C$ in the Kummer variety of $A$, that is $\mbox{Km}(A)=A/\pm$, and so $C$ contains $2g+2$ 2-torsion points of $A$. Let $2_A$ be the morphism multiplication by 2 on $A$, and let $C'=2_A(C)$. This is a (possibly singular) curve on $A$, and the image of $C'$ in $\mbox{Km}(A)$ is also rational (since it is the image of $2_{\mbox{Km}(A)}$ restricted to the projection of $C$ in $\mbox{Km}(A)$). If we take the normalization of $C'$, let's say $\pi:N\to C'$, then we get a map $C\to N$, that lifts $2_A:C\to C'$. Now, $N$ is hyperelliptic and if $g\geq 2$, we must have that $g_N$ (the genus of $N$) is less than $g$, unless the map $C\to N$ is an isomorphism. The map can never be an isomorphism, given that the $2g+2$ torsion points on $C$ all go to 0 in $C'$, and above each point in $C'$ there are at most 2 points in $N$.
One very false implication of this reasoning is that if $C$ is of genus 2, for example, and $A$ is the jacobian of $C$, then $N$ would have to be an elliptic curve! This would mean that every 2-dimensional abelian variety is isogenous to the product of elliptic curves, which is very very false!
What's wrong with this argument?