Suppose that $\preimage{T}{\vect{v}}$ is a subspace of $U$. Then $\preimage{T}{\vect{v}}$ is nonempty so we can apply Theorem KPI, and assert the existence of a vector $\vect{u}\in\preimage{T}{\vect{v}}$ so that $\preimage{T}{\vect{v}}=\vect{u}+\krn{T}$. Furthermore, if $\preimage{T}{\vect{v}}$ is a subspace, then $\zerovector\in\preimage{T}{\vect{v}}$, so there exists a vector $\vect{z}\in\krn{T}$ such that $\zerovector=\vect{u}+\vect{z}$. Now
\begin{align*}
\vect{v}&=\vect{v}+\zerovector&&\knowl{./knowls/property.Z.knowl}{\text{Property Z}}\\
&=\vect{v}+\lt{T}{\vect{z}}&&\knowl{./knowls/definition.KLT.knowl}{\text{Definition KLT}}\\
&=\lt{T}{\vect{u}}+\lt{T}{\vect{z}}&&\knowl{./knowls/definition.PI.knowl}{\text{Definition PI}}\\
&=\lt{T}{\vect{u}+\vect{z}}&&\knowl{./knowls/definition.LT.knowl}{\text{Definition LT}}\\
&=\lt{T}{\zerovector}\\
&=\zerovector&&\knowl{./knowls/theorem.LTTZZ.knowl}{\text{Theorem LTTZZ}}\\
\end{align*}
So we our hypothesis that the preimage is a subspace has lead to the conclusion that $\vect{v}$ could only be one vector, the zero vector. We still need to verify that $\preimage{T}{\zerovector}$ is indeed a subspace, but since $\preimage{T}{\zerovector}=\krn{T}$ this is just Theorem KLTS.