Proof of Theorem ILTD

Suppose to the contrary that $m=\dimension{U}>\dimension{V}=t$. Let $B$ be a basis of $U$, which will then contain $m$ vectors. Apply $T$ to each element of $B$ to form a set $C$ that is a subset of $V$. By Theorem ILTB, $C$ is linearly independent and therefore must contain $m$ distinct vectors. So we have found a set of $m$ linearly independent vectors in $V$, a vector space of dimension $t$, with $m>t$. However, this contradicts Theorem G, so our assumption is false and $\dimension{U}\leq\dimension{V}$.