Let $K\subset L$ be a field extension and $A, B\subset L$ be $K$-subspaces of $L$ of finite positive dimensions. Assume further that for every $a, b \in L$ and every nontrivial proper finite intermediate subfield $K\subset M\subsetneq L$, we have $$ \dim_{K}(aM\cap A) + \dim_{K} (bM\cap B)\leq [M:K]+1. $$ Then does the following hold? $$ \dim_{K}\langle AB\rangle\geq\dim_{K} A+\dim_{K} B-1, $$ where $AB=\{ab\mid a\in A, b\in B\}$ and $\langle AB\rangle$ is the $K$-subspace of $L$ spanned by $AB$.
Intuition: For every proper finite intermediate field $K \subset M \subsetneq L$, and for any $a, b \in L$, the assumption $\dim_K(aM \cap A) + \dim_K(bM \cap B) \leq [M : K] + 1$ restricts the "concentration" of $A$ and $B$ inside proper subfields. Roughly speaking, it avoids the case where a large part of both $A$ and $B$ live inside a small field extension — a situation that could lead to a "small" product span $\langle AB \rangle$.
Remark 1: Note that a similar result in a group setting is true. (Replace $K\subset L$ with an abelian group $G$, subspaces with subsets and dimension with cardinality.) That is, we have the following result:
Theorem: Let $G$ be an abelian group and $A$ and $B$ be nonempty finite subsets of $G$. Assume that for every nontrivial finite subgroup $H$ of $G$ and all elements $a,b\in G$, we have $|aH\cap A|+|bH\cap B|\leq|H|+1$. Then we have $|AB|\geq|A|+|B|-1$.
Remark 2: In the linear setting we have the following theorem in this paper: Let $K\subset L$ be a field extension and $A\subset L$ be a $K$-subspace of $L$ of finite positive dimension. Then either for every finite-dimensional subspace $B$ of $L$, we have \begin{align*} \dim_K\langle AB\rangle\geq\dim_K A+\dim_K B-1, \end{align*}
or there exists an intermediate field $K\subset M\subset L$, such that for every finite-dimensional subspace $B$ of $L$ satisfying \begin{align*} \dim_K\langle AB\rangle < \dim_K A+\dim_K B -1, \end{align*} we have $\langle AB\rangle M=\langle AB\rangle$.
Remark 3: Another related theorem from this paper: Let $K\subset L$ be a field extension in which every algebraic element of $L$ is separable over $K$. Let $A,B\subset L$ be nonzero finite dimensional $K$-subspaces of $L$ and $M$ be the stabilizer of $\langle AB\rangle$, i.e., $M=\{x\in L| x\langle AB\rangle\subset \langle AB\rangle\}$. Then \begin{align*} \dim_K\langle AB\rangle\geq \dim_K A+\dim_K B-\dim_K M. \end{align*}
Remark 4: For every nonzero finite-dimensional $K$-subspace $X$ of $L$, we denote by \begin{align*} \partial_A(X)=\dim_K\langle XA\rangle-\dim_{K} X. \end{align*} Let $\psi$ be the set of nonzero finite-dimensional $K$-subspaces $X$ of $L$ such that $\langle XA\rangle\neq L$. If $\psi$ is nonempty, we define the 1st-connectivity of $A$ by \begin{align*} \kappa(A)=\min_{X\in\psi}\partial_A(X). \end{align*} If $\psi$ is empty, we set $\kappa(A)\neq-\infty$. We define a 1st-fragment of $A$ to be an $M\in\psi$ with $\partial_A(M)=\kappa(A)$. A 1st-fragment with minimum dimension is called a 1-atom.
We now have the following theorem from this paper:
Theorem: Let $K\subset L$ be a field extension. Let $S$ be a $K$-subspace of $L$ containing $1$. Let $A$ be the 1-atom of $S$. If there exists a $K$-subspace $T$ of $L$ for which $\dim_{K}\langle ST\rangle<\dim_{K} S+\dim_{K}T-1<\dim_{K} L$, then $A$ is a subfield of $L$ properly containing $K$.
Remark 5: A result on large subspaces from this paper: Let $K \subseteq L$ be a field extension, with $\dim_K L$ finite. Let $A, B$ be non-zero subspaces of $L$ satisfying $\dim_K A + \dim_K B \geq \dim_K L$. Then $\langle AB \rangle = L$.
Remark 6: Let $L \subset L$ be a field extension. Define $\mu_{K,L}(r,s) = \min \dim_K <AB>$, where the minimum is taken over all $K$-subspaces $A, B$ of $L$ satisfying $\dim_K A = r$, $\dim_K B = s$. Define $\kappa_{K,L}(r,s) = \min_{h}(\lceil\frac{r}{h}\rceil+\lceil\frac{s}{h}\rceil-1)h $, where $h = [H : K]$ runs over the set of $K$-dimensions of all finite-dimensional intermediate fields $K \subset H \subset L$. It is proved in this paper that if $K \subset L$ is a field extension in which every algebraic element of $L$ is separable over $K$, then, for all positive integers $r, s \leq \dim_K L$, we have $\mu_{K,L}(r,s) = \kappa_{K,L}(r,s)$.