I read somewhere that
The $n-r$ special solutions of a $m \times n$ matrix with rank $r$ form a basis for its null-space.
If we consider the general RREF for the given matrix, it has the form:
\begin{pmatrix} I & F \\ 0 & 0 \end{pmatrix}
where $I$ is the $r \times r$ identity matrix and $F$ is a $r \times (n-r)$ matrix. The special solutions can be grouped into a solution matrix of the form
\begin{pmatrix} -F \\ I \end{pmatrix}
I see nothing in the structure of the solution that would suggest these from a basis. If it was only $I$, that would have made sense. But $F$ is completely general.
So,
Why do special solutions of $Ax=0$ form a basis for null-space of $A$ ?
Note that $F$ is not completely general. It is uniquely determined from your matrix (since the RREF is uniquely determined). Also note that through block matrix multiplication, that each column of your "special matrix" (we'll call this matrix $A$ from now on) is indeed a vector in the nullspace: $$\begin{pmatrix} I_r & F\\ 0 & 0 \end{pmatrix}\begin{pmatrix} -F \\ I_{n-r}\end{pmatrix} = \begin{pmatrix} -I_rF + FI_{n-r} \\ 0 \end{pmatrix} = \begin{pmatrix}0 \\ 0\end{pmatrix}$$ I have added the sizes of the identities for clarity.
Also note that the columns of $A$ are linearly independent. This is trivially so because $I_{n-r}$ itself is linearly independent. So we know the columns of $A$ are $n-r$ independent vectors in the nullspace, which necessarily forms a basis for the nullspace since we know from the rank-nullity theorem that the nullspace is of dimension $n-r$.