Here is a more explicit step-by-step version of the proof quoted in the question.
Proof
Let $A\in\mathcal{M}_{m\times n}(\mathbb{K})$, where $\mathbb{K}$ is any field. Let $\{\boldsymbol{u^1},\ldots, \boldsymbol{u^k} \}$ be a basis of the column space of $A$, where $k\in\{1,\ldots,n\}$. $\boldsymbol{u^i}$ is a vector in $\mathbb{K}^m$ (the vector space of $m$-tuples with entries in $\mathbb{K}$) for all $i\in\{1,\ldots,k\}$. Thus each column in $A$ can be expressed as a linear combination of $\boldsymbol{u^1},\ldots,\boldsymbol{u^k}$. That is, for every $j\in\{1,\ldots,n\}$ there exist unique coefficients $\lambda_{1j},\ldots,\lambda_{nj}\in\mathbb{K}$ such that
$$
\forall j\in\{1,\ldots,n\}\qquad \boldsymbol{v^j} = \sum_{\ell=1}^{k} \lambda_{\ell j} \boldsymbol{u^\ell}\,,
$$
where $\boldsymbol{v^j}$ denotes the $j$-th column in $A$.
Let $B\in\mathcal{M}_{m\times k}(\mathbb{K})$ be the matrix with such vectors as columns:
$$
B =
\begin{pmatrix}
\vert & & \vert \\
\boldsymbol{u^1} & \cdots & \boldsymbol{u^k} \\
\vert & & \vert
\end{pmatrix}\,.
$$
Let $[s]$ denote $\{1,\ldots,s\}$ for all $s\in\mathbb{N}$. Let $C\in\mathcal{M}_{k\times n}(\mathbb{K})$ be the matrix with the aforementioned coefficients:
$$
C = (\lambda_{ij})_{(i,j)\in[k]\times[n]} =
\begin{pmatrix}
\lambda_{11} & \cdots & \lambda_{1n} \\
\vdots & \ddots & \vdots \\
\lambda_{k1} & \cdots & \lambda_{kn}
\end{pmatrix}\,.
$$
Now consider the matrix product $BC\in\mathcal{M}_{m\times n}(\mathbb{K})$. Let $(bc)_{ij}$, $b_{ij}$ and $c_{ij}$ denote the $(i,j)$-th element of $BC$, $B$ and $C$ respectively. By definition of matrix product,
\begin{equation}\tag{1}\label{foo}
(bc)_{ij} = \sum_{\ell=1}^{k} b_{i\ell}c_{\ell j} \qquad\forall (i,j)\in[m]\times[n]\,.
\end{equation}
Let us consider the $j$-th column of $BC$ for an arbitrary $j\in[n]$. Let $v^\ell_i$ denote the $i$-th component of $\boldsymbol{u^\ell}$ for all $\ell\in[k]$ and for all $i\in[m]$.
$$
\begin{multline*}
\left((bc)_{ij}\right)_{i\in[m]} = \left( \sum_{\ell=1}^{k} b_{i\ell}c_{\ell j} \right)_{i\in[m]} =
\begin{pmatrix}
\sum_{\ell=1}^{k} b_{1\ell}c_{\ell j} \\
\vdots \\
\sum_{\ell=1}^{k} b_{m\ell}c_{\ell j}
\end{pmatrix} = \\
%
=
\begin{pmatrix}
\sum_{\ell=1}^{k} v^\ell_1\cdot\lambda_{\ell j} \\
\vdots \\
\sum_{\ell=1}^{k} v^\ell_m\cdot\lambda_{\ell j}
\end{pmatrix} = \sum_{\ell=1}^k \lambda_{\ell j} %
\begin{pmatrix}
v^\ell_1 \\
\vdots \\
v^\ell_m
\end{pmatrix} = \sum_{\ell=1}^{k} \lambda_{\ell j}\boldsymbol{u^\ell} = \boldsymbol{v^j}\,.
\end{multline*}
$$
Thus, the columns of $BC$ are the columns of $A$. Ergo, $A=BC$.
On the other hand, let us consider the $i$-th row of $A$, denoted by $\boldsymbol{r^i}$. That is,
$$
\boldsymbol{r^i} = (a_{ij})_{j\in[n]} \qquad \forall i\in[m]\,.
$$
Again, by the definition of matrix multiplication,
$$
a_{ij} = \sum_{\ell=1}^{k} b_{i\ell}c_{\ell j} \qquad\forall (i,j)\in[m]\times[n]
$$
(this is the same equation found in eq. \eqref{foo}). Thus,
$$
\begin{multline}\tag{2}\label{rows}
\boldsymbol{r^i} = \left(\sum_{\ell=1}^{k} b_{i\ell}c_{\ell j} \right)_{j\in[n]} =
\begin{pmatrix}
\sum_{\ell=1}^{k} b_{i\ell}c_{\ell 1} & \cdots & \sum_{\ell=1}^{k} b_{i\ell}c_{\ell n}
\end{pmatrix} = \\
=
\begin{pmatrix}
\sum_{\ell=1}^{k} v^\ell_i\cdot\lambda_{\ell 1} & \cdots & \sum_{\ell=1}^{k} v^\ell_i\cdot\lambda_{\ell n}\,.
\end{pmatrix}
\end{multline}
$$
Now, let $\boldsymbol{\Lambda^\ell}$ be the $\ell$-th row of $C$ for all $\ell\in[k]$, as a row vector:
$$
\begin{equation*}
\boldsymbol{\Lambda^\ell} =
\begin{pmatrix}
\lambda_{\ell 1} & \cdots & \lambda_{\ell n}\,.
\end{pmatrix}
\end{equation*}
$$
Thus, with the same notation as before, ${\Lambda^\ell_i} = \lambda_{\ell i}$ for all $i\in[n]$. Also, let $\mu_{i \ell}$ denote $v^\ell_i$ for all $i\in[m]$ and for all $\ell\in[k]$ — this is merely a change of notation to remark the fact that $v^\ell_i$ can be seen as "coefficients". Thus, continuing to develop equation \eqref{rows}, we get
$$
\begin{multline*}
\boldsymbol{r^i} =
\begin{pmatrix}
\sum_{\ell=1}^{k} \mu_{i\ell}\cdot\Lambda^\ell_1 & \cdots & \sum_{\ell=1}^{k} \mu_{i\ell}\cdot\Lambda^\ell_n
\end{pmatrix} = \\
=
\sum_{\ell=1}^k \mu_{i\ell}
\begin{pmatrix}
\Lambda^\ell_1 & \cdots & \Lambda^\ell_n
\end{pmatrix} =
\sum_{\ell=1}^k \mu_{i\ell} \boldsymbol{\Lambda^\ell}\,.
\end{multline*}
$$
Therefore, the rows of $A$ (i.e., $\boldsymbol{r^i}$) are linear combinations of the rows of $C$ (i.e., $\boldsymbol{\Lambda^\ell}$). Thus, we necessarily have
$$
\mathrm{rowsp}\ A \subseteq \mathrm{rowsp}\ C \ \implies\ \dim (\mathrm{rowsp}\ A) \le \dim (\mathrm{rowsp}\ C)\,.
$$
Since $C$ has $k$ rows, its row space can have at most dimension $k$, which is the dimension of $\mathrm{colsp}\ A$ (by hypothesis):
$$
\dim(\mathrm{rowsp}\ C) \le k = \dim(\mathrm{colsp}\ A)\,.
$$
Combining both inequalities, we have
$$
\dim (\mathrm{rowsp}\ A) \le \dim (\mathrm{colsp}\ A)\,.
$$
Applying this whole argument again on $A^\mathrm{t}$,
$$
\dim (\mathrm{rowsp}\ A^\mathrm{t}) \le \dim (\mathrm{colsp}\ A^\mathrm{t}) \iff \dim (\mathrm{colsp}\ A) \le \dim (\mathrm{rowsp}\ A)\,.
$$
Since we have both $\dim (\mathrm{rowsp}\ A) \le \dim (\mathrm{colsp}\ A)$ and $\dim (\mathrm{colsp}\ A) \le \dim (\mathrm{rowsp}\ A)$, we conclude that
$$
\dim (\mathrm{colsp}\ A) = \dim (\mathrm{rowsp}\ A)\,. \quad \square
$$
