If every vector is an eigenvector, the operator must be a scalar multiple of the identity operator? I am posed with the following question:

Suppose that $T\in \mathcal{L}(V)$, where $V$ is a finite-dimensional vector space,  is such that every vector in $V$ is an eigenvector of $T$. Prove that T is a scalar multiple of the identity function.


My attempt goes as follows:
If $T$ is a scalar multiple of the identity function, then $Tv=av$ for all $v\in V$, where $a$ doesn't depend on $v$. We will start off assuming that that may not necessarily be true, and work our way to the result.
From the problem statement, we know that $Tv=a_j v$, where $a_j$ may depend on the choice of $v$. To show that it doesn't, consider two non-zero vectors $v_1$ and $v_2$ both in $V$ (It would be pointless to consider zero vectors). Consider $T(v_1+v_2)$. Also, let us first consider the case where $v_2$ is not a scalar multiple of $v_1$ (so $v_1$ & $v_2$ form a linearly independent set). On one hand we have
$$\begin{align*}
T(v_1+v_2)&=\alpha (v_1+v_2)\\
&= \alpha v_1 + \alpha v_2
\end{align*}$$
Which is true because of the the assumption that every vector is an eigenvector. On the other hand, we have
$$\begin{align*}
T(v_1+v_2)&=T(v_1)+T(v_2)\\
&=a_1 v_1 + a_2 v_2
\end{align*}$$
So we are left with the following equality
$$a_1 v_1 + a_2 v_2=\alpha v_1 +  \alpha v_2$$
$$\implies (a_1 - \alpha)v_1 + (a_2 - \alpha)v_2 = 0$$
Because $v_1$ & $v_2$ are linearly independent, $a_1=a_2=\alpha$.
Now consider $\beta v$, a scalar multiple of $v$. 
$$\begin{align*}
T(\beta v)&= \beta T(v)\\
&= \beta (a v)\\
&= a (\beta v)
\end{align*}
$$
Because every vector in $V$ that is not equal to an arbitrary vector $v$ is either a scalar multiple of it or the sum of $v$ and some other vector in $V$, $Tv=av$ for all $v\in V$.

Is my prove valid? Please criticize my proof holistically. 
P.S: Is the same proof valid for infinite-dimensional vector spaces? I ask because such truth would lie in the validity of the last paragraph, and I don't know if it is true in infinite-dimensional vector spaces.
 A: Here's as holistic approach as I've got to the present problem:
This very same question actually came up in a research project on differential equations I last worked on a couple of years ago, so it is something with which I am well familiar.  And my proof was essentially the same as OP Arturo Don Juan's, so I believe I can affirm the correctness of his argument with confidence.
Just for completeness, here's the way I wrote it up (from my research notes):
Let $\Bbb F$ be any field, and let $V$ be any vector space over $\Bbb F$ (note we don't require $\dim V < \infty$.).  If $T \in \mathcal L(V)$ is such that every non-zero $v \in V$ is an eigenvector, then there is an $\alpha \in \Bbb F$ such that $T(v) = \alpha v$ for all $v \in V$.
Proof:  The hypothesis that every (non-zero) $v \in V$ is an eigenvector of $T$ may be expressed by the equation
$Tv = \phi(v) v, \tag{1}$
where $\phi(v) \in \Bbb F$ for every $v$; that is $\phi:V \to \Bbb F$ is a function taking vectors in $V$ to scalars in $\Bbb F$.  If $u \in V$ is related to $v \in V$ by the equation $u = \beta v$ for some $\beta \in \Bbb F$, we have
$\phi(u)u = \phi(\beta v)(\beta v) = T(\beta v) = \beta T(v) = \beta \phi(v) v = \phi(v) \beta v, \tag{2}$
or in short form,
$\phi(\beta v)(\beta v) = \phi(v) \beta v; \tag{3}$
when $\beta v \ne 0$, (3) shows that
$\phi(\beta v) = \phi(v); \tag{4}$
i.e., $\phi(v)$ only depends on the one-dimensional subspace $v$ generates; on $\text{span}\{v\}$; this implies that, for $v_1, v_2$ linearly dependent, $\phi(v_1) = \phi(v_2)$.  In the event that $v_1, v_2$ are linearly independent, we may write in manner analogous to our OP
$T(v_1 + v_2) = \phi(v_1 + v_2) (v_1 + v_2) = \phi(v_1 + v_2) v_1 + \phi(v_1 + v_2) v_2 \tag{5}$
and
$T(v_1 + v_2) = T(v_1) + T(v_2) = \phi(v_1)v_1 + \phi(v_2)v_2; \tag{6}$
combining (5) and (6):
$\phi(v_1 + v_2) v_1 + \phi(v_1 + v_2) v_2 =  \phi(v_1)v_1 + \phi(v_2)v_2; \tag{7}$
by the linear independence of $v_1$ and $v_2$, we see from (7) that
$\phi(v_1) = \phi(v_1 + v_2) = \phi(v_2); \tag{8}$
since $v_1, v_2$ are an arbitrary pair of linearly independent vectors, this result combined with (3)-(4) shows that $\phi(v)$ is a constant; taking $\alpha = \phi(v)$ for any $v \in V$ then yields the complete solution.  QED.
