$\textbf{Notation:}$
$\mathcal{A}_r(F)$ is the space of $r$-forms alternating on $F$.
$A^*$ is the pullback of a linear transformation $A$.
$\textbf{Question:}$ The rank of a linear transformation $A: E \longrightarrow F$ is the largest integer $r$ such that the pullback of $A$, i.e., $A^*: \mathcal{A}_r(F) \longrightarrow \mathcal{A}_r(E)$ is different to zero.
$\textbf{My attempt:}$
Let be $\dim F := n$ and the rank($A$) := $m$. We know that every set of $\{ v_i \in E \}$ with at least $m + 1$ elements is linearly dependent, because rank($A$) $ = m$, then given $\omega \in \mathcal{A}_r(F)$ , $v_1, \cdots, v_{m+1} \in E$ and assuming without loss of generality that $v_2 = k v_1$, we know that
$\omega(v_1, v_2, \cdots,v_{m+1}) = \omega(v_1, kv_1,\cdots,v_{m+1}) = k\omega(v_1, v_1, \cdots, v_n) = k 0 = 0$,
then $r < m+1$ by the definition of $r$ given in the statement of the question. We know too that every set of $\{ v_i \in E \}$ with at most $m$ elements is linearly independent, because rank($A$) $ = m$, then given $\omega \in \mathcal{A}_s(F)$ , $v_1, \cdots, v_{s} \in E$ with $s < m$, $\omega(v_1, v_2, \cdots,v_s) \neq 0$ and follows the definition of $r$ that $r = m. \ \square$
I would like to know if my attempt is correct. Thanks in advance!
There's a nicely written proof of this question in: Relation between rank of a linear transformation and pullback ($r$-linear forms.)
As for your answer, I think the general idea is correct and I would only like to add:
1 - After proving that $r < m + 1$ you say:
This doesn't always hold (just take a set $\{v_i \in E \}$ such that $v_1 = v_2$). But I do believe that we can choose a convenient set $\{v_i \in E \}$ such that this holds and this is enough for our purposes here.
2 - You also say:
Again, here it is not necessary that what you say holds for every $\omega$. It is enough to find at least one form $\omega$ who does not vanish to conclude what you want (check the link to see how to find a possible $\omega$ that fulfills the conditions we want).