Let $K/F$ be a field extension and $f \in F[x]$ with $\deg{(f)} \geq 1$. Furthermore, for a subset $L$ with $F\subseteq L \subseteq K$, let $p(L)$ denote the property that $f$ can be factored completely into linear factors in $L[x]$. Now consider the set $$\mathcal{L}:=\bigcap\{F\subseteq L \subseteq K \mid p(L) \ \text{holds}\} $$ Question: If property $p(K)$ holds, does it follow that $p(\mathcal{L})$ holds?
In words: If $f$ can be factored completely into linear factors in $K[x]$, does it follow that $f$ can be factored completely into linear factors in $\mathcal{L}[x]$?
Yes, that is true. Let $\alpha$ be any root of $f$ in $K$. As $\alpha$ is contained in every $L$ such that $p(L)$ holds we have $\alpha\in\mathcal{L}$. Therefore $\mathcal{L}$ contains all roots of $f$ in $K$ and $f$ splits over $K$, so $f$ will also split over $\mathcal L$ since all the linear factors over $K$ are actually defined over $\mathcal L$.