I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
The author assumed that $V$ is finite-dimensional in Exercise 5.A.28.
But I don't think that this assumption is necessary for this exercise.
The author is very careful and the author doesn't usually assume that $V$ is finite-dimensional if not necessary.
Is the assumption that $V$ is finite-dimensional really necessary for Exercise 5.A.28?
Exercise 5.A.26:
Suppose $T\in\mathcal{L}(V)$ is such that every nonzero vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator.
Proof:
Let $v,w$ be arbitrary nonzero elements of $V$.
Since $v$ and $w$ are eigenvectors of $T$ by assumption, there exist $\lambda_1,\lambda_2\in\mathbb{F}$ such that $T(v)=\lambda_1 v$ and $T(w)=\lambda_2 w$.
If $v+w\neq 0$, then there exists $\lambda_3\in\mathbb{F}$ such that $T(v+w)=\lambda_3(v+w)$.
If $v+w=0$, then $T(v+w)=T(0)=0=0(v+w)$.
So, there exists $\lambda_3\in\mathbb{F}$ such that $T(v+w)=\lambda_3(v+w)$.
Since $T(v+w)=T(v)+T(w)$, the following equality holds:
$\lambda_3(v+w)=\lambda_1 v + \lambda_2 w$.
So, $(\lambda_3-\lambda_1)v + (\lambda_3-\lambda_2)w=0$.
If $v$ and $w$ are linearly independent, then $\lambda_1=\lambda_2=\lambda_3$.
If $v$ and $w$ are linearly dependent, then $\lambda_3-\lambda_1\neq 0$ and $\lambda_3-\lambda_2\neq 0$, so we can write $w=av$ for some $a\in\mathbb{F}$.
$T(w)=T(av)=aT(v)=a\lambda_1 v=\lambda_1(av)=\lambda_1 w$ and $T(w)=\lambda_2 w$.
Since $w\neq 0$, $\lambda_1=\lambda_2$.
So, there exists $\lambda\in\mathbb{F}$ such that $T(v)=\lambda v$ for any $v\neq 0$.
Of course, $T(0)=0=\lambda 0$.
So, there exists $\lambda\in\mathbb{F}$ such that $T(v)=\lambda v$ for any $v$.
Since $\lambda v=\lambda I(v)$, $T=\lambda I$.
Exercise 5.A.28:
Suppose $V$ is finite-dimensional with $\dim V\geq 3$ and $T\in\mathcal{L}(V)$ is such that every $2$-dimensional subspace of $V$ is invariant under $T$. Prove that $T$ is a scalar multiple of the identity operator.
Proof:
Let $v$ be an arbitrary nonzero element of $V$.
Since $\dim V\geq 3$, there exist $v_1,v_2\in V$ such that $v,v_1,v_2$ are linearly independent.
By assumption $\operatorname{span}(v,v_1)$ and $\operatorname{span} (v,v_2)$ are invariant subspaces of $T$.
So, $\operatorname{span}(v,v_1)\cap\operatorname{span} (v,v_2)=\operatorname{span} (v)$ is also an invariant subspace of $T$.
So, $T(v)=\lambda v$ for some $\lambda\in\mathbb{F}$.
So, $v$ is an eigenvector of $T$.
So, by Exercise 5.A.26, $T$ is a scalar multiple of the identity operator.
We don't use the assumption that $V$ is finite-dimensional.