I was wondering if this reasoning could be considered a proof:
Let $A$ be a matrix $(m \times n)$ with $m$ rows and $n$ columns. Now we consider that there are at least $k$ rows vectors that are linearly independent, and that the span of $(v_1,v_2,...,v_k)$ row vectors includes the space of columns. So the dimension of columns is less or equal of the dimension of rows $\text{span}(v_1,v_2,...,v_k)$.
By applying the same reasononing to columns, let the columns have $r$ linearly independent vectors $(v_1,v_2,...,v_r)$ the span of columns $\supseteq$ space of rows, and therefore the dimension of rows is less or equal to the dimension of span of columns vectors $(v_1,v_2,...,v_r)$ therefore to respect both of equations dimension of column is equal of the dimension of rows.
My reasoning is probably wrong but I would like to get an advice :).
Here's a $1 \times 3$ matrix: $$ A = \pmatrix{1 & 3 & 2} $$ of row-rank $1$. Your argument says that if I look at one linearly independent row (that'd be row 1, of course), it "includes the space of columns".
Now the column-space of this matrix is just $\Bbb R$, so that claim doesn't seem to make sense.
So I'd say that your proof is flawed.
Broad hint: It's often wise to sanity-check things on low-dimensional examples.