If $E,F$ are finite fields and $F\subseteq E,$ why is $E$ a finite-dimensional vector space over $F$? I understand that if $E$ and $F$ are each finite and $E$ is a vector space over $F$, then $E$ must be a finite-dimensional vector space over $F$. However, my question is: why does $F\subseteq E$ imply that $E$ is a vector space over $F$?
 A: If $E$ is an extension of the field $F$, then $E$ is a vector space over $F$.
If $E$ is finite, it obviously has a finite spanning set as vector space over $F$, so it's finite dimensional.
Conversely, if $E$ is a finite extension of $F$, then it's finite because if $e_1,e_2,\dots,e_n$ is a basis of $E$ over $F$, then $|E|=|F|^n$.
A: Every field extension $L/K$ gives rise to a $K$-vector space in secret. 
This $K$-vector space is $L$ itself. We can definitely add things in $L$ since it is a field. We can also "scalar multiply" by $K$...just consider it as ordinary multiplication in the field (which is fine since $K\subseteq L$).
This makes sense, for example one way to view the complex numbers $\mathbb{C}$ is as numbers of the form $a+ib$ with $a,b\in\mathbb{R}$. What this is really telling you is that the field $\mathbb{C}$ is an $\mathbb{R}$-vector space with basis $\{1,i\}$.
So this settles your original question. If $E/F$ is an extension of finite fields then clearly an $F$-basis for $E$ cannot contain infinitely elements...otherwise $E$ would be infinite itself.
