The result stated is false, so you need not bother to try and prove it.
Counterexample. Let $K$ by your field, take $V=K^3$ with coordinate functions $x,y,z$, all subspaces $A,B,C,A',B',C'$ of dimension $2$ with $A=A'$ given by the equation $x=0$, $B=B'$ given by the equation $y=0$, $C$ given by the equation $z=0$ and $C'$ given by the equation $x+y=0$. You can check that the sum of any two distinct ones of those subspaces (and a fortiori of three) is all of $V$, that the intersection of two distinct among them is of dimension $1$, and that $\dim A\cap B\cap C=0$ while $C'\supset A'\cap B'$ so that $\dim A'\cap B'\cap C'=\dim A'\cap B'=1$.
Added, to answer the additional question in the comment. This example can also be used to show that given a finite set of vectors and the dimensions of the spans of all finite subsets, one cannot in general deduce the dimensions of the intersections of three or more of those spans (for the intersection of two it is possible, using $\dim(A\cap B)=\dim A+\dim B-\dim(A+B)$). It suffices to view each of the planes $A=A'$, $B=B'$, $C$, and $C'$ as the span of two specific vectors, in such a way that no three among the eight vectors used are ever coplanar (linearly dependent). Once this is achieved, the dimension of the span of any subset of at most three of those vectors is equal to their number, and the dimension of the span of any larger subset is of course $3$; nevertheless $\dim A\cap B\cap C\neq\dim A'\cap B'\cap C'$ remains true, showing that this dimension cannot be computed from the subset-span-dimension data. To see that a proper choice of $8$ vectors can be achieved, one can avoid choosing any vector lying in the intersection of two of the four planes $A,B,C,C'$; then the requirement to avoid choosing a vector coplanar with two previously chosen ones forbids a finite set of planes, but never the plane for which one is choosing a generator, so a proper choice is always possible.