= Solution
Take canonical kinetic terms, gauge coupling $g>0$, and a nonzero charge $q$. Define $e=gq$ and choose the convention in which the <Fayet–Iliopoulos term> contributes $\xi D$ to the <Lagrangian>. With one charged <chiral superfield>, gauge invariance permits no nonconstant polynomial <superpotential>; an irrelevant global constant gives $F=0$. The real <auxiliary field> part is
$$
\mathcal L_D=\frac12D^2+D(\xi+e|\phi|^2).
$$
Its equation of motion is $D=-(\xi+e|\phi|^2)$. Eliminating it gives the <D-term scalar potential>
$$
\boxed{V_D=\frac12(\xi+e|\phi|^2)^2.}
$$
The <gaugino> transformation has the form $\delta\lambda=i\epsilon D+$ a field-strength term, with a convention-dependent overall phase. In a translationally invariant vacuum with zero field strength, a nonzero $\langle D\rangle$ therefore gives an inhomogeneous shift of $\lambda$: no nonzero <supersymmetry> parameter leaves the vacuum invariant. The massless <Goldstino> is consequently the <gaugino>, not the matter fermion, since $F=0$.
Minimize the potential over $u=|\phi|^2\geq0$. If $q\xi<0$, the allowed value $u=-\xi/e$ sets $D=0$, so <supersymmetry> is unbroken while the internal gauge symmetry is Higgsed. If $\xi=0$, the origin also has $D=0$. For the broken branch, $q\xi>0$ and the formal zero of $D$ would require negative $u$. The actual minimum is
$$
\boxed{\langle\phi\rangle=0,\qquad\langle D\rangle=-\xi,
\qquad V_{\min}=\frac12\xi^2>0,
\qquad q\xi>0.}
$$
This is <single charged-field D-term breaking>. Nonzero FI parameter alone does not force breaking if a scalar can cancel it. The degenerate case $q=0$ is different: any $\xi\ne0$ breaks <supersymmetry> in the vector sector while the neutral scalar remains flat; the sign test above assumes an actually charged field.
Expand around the broken vacuum:
$$
V_D=\frac12\xi^2+e\xi|\phi|^2+\frac12e^2|\phi|^4.
$$
For $\phi=(\phi_R+i\phi_I)/\sqrt2$, both canonically normalized real scalars have squared mass $e\xi=gq\xi>0$. The matter <Weyl spinor> has no <superpotential> mass. The gauge boson remains massless because $\langle\phi\rangle=0$. The gauge Yukawa coupling, proportional to $\phi^*\lambda\psi$, gives no fermion mixing or mass at this vacuum, so the <gaugino> is also massless. The tree-level spectrum is therefore
$$
\begin{array}{c|c}
\text{field}&\text{squared mass}\\\hline
\phi_R,\phi_I&gq\xi\\
\psi&0\\
A_\mu&0\\
\lambda\ \text{(Goldstino)}&0
\end{array}
$$
and the chiral-multiplet splitting is
$$
\boxed{m_{\mathrm{scalar}}^2-m_{\mathrm{fermion}}^2=gq\xi,\qquad
m_{\mathrm{scalar}}=\sqrt{gq\xi}.}
$$
The vector and its <gaugino> happen to remain degenerate at zero mass even though <supersymmetry> is broken. If the sign convention for the FI contribution is reversed, the sign criterion changes accordingly; the invariant condition is that the scalar condensate cannot cancel the auxiliary source. This is the classical toy-model spectrum; a quantum completion of the single charged-fermion theory also has to cancel its <gauge anomalies>.
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