I am required to prove the following Claim. Is my argument valid.
Theorem. If $V$ and $W$ are vector spaces and $T$ and $U$ are non-zero Linear transformations from $V$ to $W$ such that $\operatorname{range}T\cap\operatorname{range}U = \{0\}$. then the set $\{U,V\}$ is linearly independent in $\mathcal{L}(V,W)$.
Proof. Assume on the contrary that the set $\{U,V\}$ is linearly dependent that is for some $c\in\mathbf{F}$ either $T = cU$ or $U = cT$. Let us first consider the former case.
Since $T$ and $U$ are non-zero linear maps it follows that $c\neq 0$ and that for some non-zero vector $v\in V$ we have $Tv = U(cv)\neq 0$ evidently $Tv\in\operatorname{range}T$ and since $Tv = U(cv)$ it follows that $Tv\in\operatorname{range}U$ consequently $Tv\in\operatorname{range}T\cap\operatorname{range}U$ and so $Tv = 0$ resulting in a contradiction.
We may arrive at a similar absurdity when addressing the case $U = cT$.
$\blacksquare$
The proof is good.
You could simplify it by observing that from $T=cU$, you have $T(v)=U(cv)$, for every $v\in V$. Thus $T(v)\in\operatorname{range}T\cap\operatorname{range}U$, forcing $T(v)=0$, for every $v\in V$, which is a contradiction to $T\ne0$.