If $V$ is a finite-dimensional vector space over the field $K$, and if $F$ is a subfield of $K$ such that $[K:F]$ is finite, show that $V$ is a finite-dimensional vector space over $F$ and that moreover $\dim_F V=(\dim_K V) [K:F]$.
I know that $\dim_F V=(\dim_K V) [K:F]$ can be changed into $[V:F]=[V:K][K:F]$, and it is similar to the theorem
Let $F\subseteq K\subseteq V$, then $[V:F]=[V:K][K:F]$.
But is it able to drive the relation $K\subseteq V$?
Well, as far as the final relation is concerned, it is not true that you can embed $K\subset V$ in a canonical way.
I would suggest looking at a basis $\{ v_1, v_2, \ldots, v_k\}$ of $V$ as a $K$-vector space. Here $\dim_K(V) = k$.
Since $K$ is an $F$-vector space as well, it has a basis, say $\{a_1, a_2, \ldots , a_n\}$.
Now show that $\{a_1v_1, \ldots, a_nv_1, a_1v_2, \ldots, a_nv_k\}$ gives a basis for $V$ as an $F$-vector space. Since the $a_i$ form a basis over $F$, they are linearly independent over $F$. I leave the details that this forces this set to be $F$-linearly independent to you.
Then, this basis has $\dim_F(V) = k\cdot n = \dim_K(V)[K:F]$ elements.