Is my attempt at the following Proof Correct?
We also have the following result.
$(25)$ Given that $T\in\mathcal{L}(V)$ and $u$ and $v$ are eigenvectors of $T$ such that $u+v$ is also an eigenvector then $u$ and $v$ correspond to the the same eigenvalue.
Theorem. Given that $T\in\mathcal{L}(V)$ such that every non-zero vector in $V$ is an eigenvector of $T$ show that $T$ is a scalar multiple of the linear opearator.
Proof. Let $u$ be any non-zero vector in $V$, consequently $Tu = ku$ for some $k\in\mathbf{F}$. Now let $v$ be an arbitrary non-zero vector in $V$ and consider the following Cases.
Case-1($v\neq -u$): Since $v$ is not the additive inverse of $u$ it follows that $u+v$ is non-zero and consequently $T(u+v) = \lambda(u+v)$ for some $\lambda\in\mathbf{F}$ and thus by appealing to result $(25)$ we may deduce that $k = \lambda$.
Case-2($v = -u$): We know that $Tu = ku$ thus by appealing to the homogenity of $T$, we have $T(v) = T(-u) = T((-1)\cdot u) = (-1)\cdot Tu = (-1)\cdot ku = k\cdot(-1)\cdot u = k(-u) = kv$.
Consequently $\forall u\in V\backslash\{0\}(Tu = ku)$.
$\blacksquare$
It seems correct, but it's easier. Given $u,v\in V$, you know that $$ T(u)=\alpha u,\qquad T(v)=\beta v\qquad T(u+v)=\gamma(u+v) $$ If $u$ and $v$ are linearly independent, we have therefore $$ \alpha u+\beta v=\gamma u+\gamma v $$ which yields $\alpha=\gamma$ and $\beta=\gamma$.
If $u$ and $v$ are linearly dependent there is nothing to prove, because they surely belong to the same eigenspace.