Problem Statement: Suppose $T\in\mathcal{L}(V)$ is such that every $v\in V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator.
Proof: Let $v,w\in V$. Then by assumption there exist scalars $\alpha,\beta$ such that $T(v)=\alpha v$ and $T(w)=\beta w$. Now consider $T(v+w)$. Since $V$ is a vector space it follows that $v+w\in V$ and so there exists a scalar, $\gamma$, such that $T(v+w)=\gamma (v+w)$. But $T$ is a linear operator and so it follows that $T(v+w)=\alpha v+\beta w$ and so setting these two equations equal we see that $\alpha v+\beta w=\gamma(v+w)=\gamma v+\gamma w$. Equating like terms we get $\alpha=\gamma$ and $\beta=\gamma$ and so it follows that $\alpha=\beta=\gamma$. Thus, for any $v\in V$, $T(v)=\beta v$ for some scalar $\beta$.
Now we must consider if $v,w$ are already scalar multiples of each other. Consider: $w=\eta v$ and $T(v)=\mu v$. Then $T(w)=T(\eta v)=\eta T(v)=\eta\mu v=\mu\eta v=\mu w$. An argument similar to what was done above to show that $\alpha=\beta=\gamma$ will show that all of these scalars are in fact the same scalar.
Thus, $T$ is a scalar multiple of the identity operator.
$\Box$