Super GoldStone model and SuperHiggs theorem

I am currently studying SuperSymmetry and I have reached a problem which I have not found an answer to.

I clearly understand how the Goldstone theorem works for the boson case (without any susy) however, the way I have seen the theorem, is by working with the SSB matrix, and see what generators are broken or not.

Now, I am learning SUSY by studying notes that follow Wess and Bagers book which state that you have SuperSymmetric Spontaneous breaking when your energy is positive (which I understand why), as therefore you have to see if there are solutions for whom the scalar potential vanishes or not.

In a usual Chiral case only, you need the Spinor fields to have zero mass in order to break susy which again I understand.

My problem comes when we take a supergauge invariant vector multiplet case where in order to break susy we need to enter a Fayet-Iliopoulos term.

Now, the book only argues that because δλ will be non-zero (where λ will be the spinor field in this case), λ will be our goldstino fermion now.

I dont really understand, why having a non-zero variation of the vacuum state of λ, will result to λ being massless and therefore connecting to when a generator breaks or not.

Thanks

SUSY is spontaneously broken if SUSY variation of a vacuum state of any field $$A$$ is non-zero: $$\delta_\epsilon \langle A\rangle\neq 0.$$ But the only fields that can have non-zero SUSY variation of their vacuum states are fermions since $$\delta_\epsilon\langle\psi\rangle\sim\epsilon\langle F\rangle\\ \delta_\epsilon \langle\lambda\rangle\sim\epsilon\langle D\rangle.$$ Here $$\psi$$ is a fermion from a chiral multiplet, and $$\lambda$$ is a gaugino; $$F$$ and $$D$$ are the corresponding auxiliary fields. Thus, if $$\langle F\rangle$$ and $$\langle D\rangle$$ are zero then the vacuum is supersymmetric, if either of them is non-zero then SUSY is broken.
Considering a theory of a single U(1) gauge multiplet, if SUSY is spontaneously broken Goldstone says there must be a massless goldstino. But the only fermion is the gaugino $$\lambda$$, so it will play the role of the goldstino.