 Solution

The line bundle translation stabilizer is the closed subgroupEach difference $x_1-x_2$ lies in $K(L)$, by the equality of Weil divisors just proved. Thus the subgroup they generate lies in $K(L)$, and closedness puts its Zariski topology closure there too. Therefore $\boxed{Y\subseteq K(L)}$. The closure of a subgroup is again a subgroup: multiplication and inversion preserve its closure by continuity. In this setting it is connected as well, since $C-C$ is connected and contains zero, and the increasing finite sums of $C-C$ are connected with a common point. Its reduced closure is an abelian subvariety.