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A smooth projective curve $C\subset A$ generates $A$ by its point differences if and only if it meets every prime Weil divisor. If a prime Weil divisor avoids $C$, its restriction to $C$ has degree zero; constancy of line bundle degree in a family shows that every translate either contains $C$ or misses it. This forces invariance under $C-C$ and puts the generated subgroup inside the line bundle translation stabilizer. If this subgroup were all of $A$, the divisor would be algebraically trivial, contradicting its positive intersection with an ample line bundle. Conversely a proper closed generated subgroup admits a disjoint divisor by the pole divisor avoiding a fiber construction, after translating the fiber to contain $C$.

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.