To show that this is a subspace, we need to show that it is non-empty and closed under scalar multiplication and addition. We know it is non-empty because $T(0_{m}) = 0_{n}$, so $0_{n} \in T(U)$.
Now, suppose $c \in R$ and $v_{1}, v_{2} \in T(U)$. We need to show that $cv_{1} + v_{2} \in T(U)$.
Note that since $v_{1}, v_{2} \in T(U)$, there exists some $w_{1}, w_{2} \in R^{m}$ such that $T(w_{1}) = v_{1}, T(w_{2}) = v_{2}$. Since $R^{m}$ is a vector space, $cw_{1} + w_{2} \in R^{m}$.
Now, by the properties of a linear transformation, we know that $T(cw_{1}+w_{2}) = cT(w_{1}) + T(w_{2}) = cv_{1} + v_{2}$.
I am done, I constructed a member of $R^{m}$ that maps to $cv_{1} + v_{2}$, and thus it must be that $cv_{1} + v_{2} \in T(U)$.