This is false in general, as I have indicated in my answer to the question linked to above. Apparently my recipe was too hard to execute, so I'll do so here.
We want to define four planes in $K^3$ (where $K$ is the base field), given by equations $x=0$, $y=0$, $z=0$ and $x+y=0$ respectively, each as the span of two out of $8$ vectors $v_1,\ldots,v_8$, where no triple of these vectors are linearly dependent. This can be done (for $K=\mathbf Q$) by taking $v_j$ to be column $j$ of the following matrix $$ \begin{pmatrix} 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 & 2 &-1 &-1 \\ 1 & 2 & 1 & 2 & 0 & 0 & 5 & 6 \\ \end{pmatrix}, $$ for which one can check that all $56$ of its $3\times 3$ minors are nonzero. As a consequence the span of any $d$ distinct vectors $v_j$ is of dimension $\min(d,3)$.
Now taking $S=\{v_1,v_2,v_3,v_4,v_5,v_6\}$ and $S'=\{v_1,v_2,v_3,v_4,v_7,v_8\}$ (so $v'_i=v_i$ for $i\leq 4$ and $v'_5=v_7, v'_6=v_8$), and then $A_1=A'_1=\{v_1,v_2\}$, $A_2=A'_2=\{v_3,v_4\}$, $A_3=\{v_5,v_6\}$ and $A'_3=\{v'_5,v'_6\}=\{v_7,v_8\}$, one has $$ 0=\dim\span(A_1)\cap\span(A_2)\cap\span(A_3)\neq\dim\span(A'_1)\cap\span(A'_2)\cap \span(A'_3)=1. $$ It may be noted that an intersection of at least three subspaces is needed, since $$ \dim(A\cap B)=\dim A+\dim B-\dim(A+B). $$ Note also that although the intersection $A_1\cap A_2$ occurs on both sides, I have avoided choosing any of the $v_i$ on that line ($x=y=0$).