Intuition for the choice of vector when showing that $(\forall\psi\in H:\left<\psi\mid T\psi\right>=0)\implies T=0$ for a bounded operator $T$

Question

This answer https://math.stackexchange.com/a/372246/820472 in the post Prove that the only operator on $\mathbb{C}$ for which his inner product is zero is zero shows perfectly why for a linear mapping/bounded operator $T$ it holds that if $\left<\psi\mid T\psi\right> = 0$ for all $\psi\in H$ (here $H$ a Hilbert space), then $T = 0$. What I am missing is intuition on why one might choose $\psi = \lambda \phi + T\phi, \lambda \in \mathbb{K},$ as the vector to work with in this case. Right now the aforementioned proof seems kinda backwards to me in the sense that for the uninitiated, you first have to prove the claim to realize that $\psi = \lambda \phi + T\phi$ is the easiest vector to work with. Are there some good analogies to help solidify intuition in this particular case, or is this one of the rite of passage proofs that you either "just get" initially or get by any means (read: Google) necessary?