 Solution

The field term explicitly selects the positive $x$ direction. For $B>0$ the uniform minimizing state is ${\mathbf{S}}_j=S\hat{\mathbf{x}}$, with no remaining continuous degeneracy of its orientation. Rotations about the $x$ axis still leave the Hamiltonian operator invariant, and the selected state respects that residual symmetry.

Tilting the magnetization now costs a field energy. Accordingly the transverse spin wave acquires a gapas confirmed directly in d. The factor $S$ is present because the source couples the field as $-BS{\sum }_j{S}_j^x$ rather than $-B{\sum }_j{S}_j^x$. Gapless zero-field Goldstone modes need not survive an explicit symmetry-breaking field.