A linear map on V (over $\mathbb F$) whose null space and range (space) are both finite dimensional, means V is finite dimensional I wrote up the following "proof". I've marked bits that feels especially iffy. Naturally the solutions had a much better proof than mine :) But I would like to know how well mine holds up to scrutiny.
Problem statement:

Suppose there exists a linear map T on V (over $\mathbb F$) whose null space and range ar both finite dimensional. Prove that V is finite dimensional

My attempt:
Let $r_1,\ldots, r_n$ be a basis for $\text{range } T$ (where $\dim \text{range } T = n$)
Let $v_1, \ldots, v_n$ be vectors such that $Tv_i = r_i$ ($1 \leq i \leq n$).
Then for some vector $r \in \text{range } T$ and constants $a_1,\ldots,a_n \in \mathbb F$:
$$\begin{align}
r&=a_1 r_1 + \ldots + a_n r_n\\
r&=a_1Tv_1 + \ldots + a_n Tv_n = T(a_1 v_1 + \ldots + a_n v_n)
\end{align}
$$
We then have a $U = \text{span}(v_1,\ldots,v_n)$ which is a subspace of V and $$\text{range }T = \{Tv : v=a_1 v_1+\ldots+a_n v_n,\quad a_1,\ldots,a_n \in \mathbb F\}$$

This following part feels iffy, can we generally construct $p-q$?

Assume V is infinite dimensional.
Let $p-q \in U$, where $p\in V$, $q\notin U$. 
Then $T(p-q) = Tp - Tq$ is in $\text{range }T$ which implies $q$ can be written as a linear combination of $v_1,\ldots,v_n$. Given that $q$ may be any vector in V, this contradicts that V is infinite dimensional.
 A: Unfortunately, your attempt cannot be accepted as a proof. For one thing, you use nowhere the hypothesis that the null space is finite dimensional.
(And the final part of your argument is faulty, as explained in the comments.)
There are not many ways to show a vector space is finite dimensional; either you show a finite spanning set or you prove the space is (isomorphic to) a subspace of another one you know is finite dimensional.
In this case, finding a finite spanning set is the best way.
Let $r_1,\dots,r_n$ be a finite spanning set of the range of $T$ (a basis, if you prefer). Consider $v_1,\dots,v_n$ such that $Tv_i=r_i$.
Now, for $v\in V$, you know that $Tv=\alpha_1r_1 +...+ \alpha_nr_n$; if $v'=\alpha_1v_1+\dots+\alpha_nv_n$, then $Tv=Tv'$, so $v-v'$ belongs to the null space. If $w_1,\dots,w_m$ is a spanning set for the null space, we have $v-v'=\beta_1w_1+\dots+\beta_mw_m$; therefore
$$
v=\alpha_1v_1+\dots+\alpha_n v_n+\beta_1w_1+\dots+\beta_mw_m
$$
and, since $v$ was arbitrary, we conclude that $\{v_1,\dots,v_n,w_1,\dots,w_m\}$ is a finite spanning set for $V$.
