Suppose $E$ and $F$ are subfields of $\mathbb{R}$ with $F\subseteq E$. Prove that any two bases of $E/F$ have the same cardinality.
The definition of a basis I am using is any finite set $S\subseteq E$ that is linearly independent (i.e. no element can be expressed as linear combination of all of the others) and spans $E$ (i.e. any element of $E$ can be expressed as a linear combination of elements in $S$).
I'm assuming that $E$ is a finite extension of $F$ and that they are both subfields of $ \mathbb{R}$. I know the result is true even if these assumptions are dropped, but I'd like to prove it in this case specifically without resorting to more general results of field theory. How can I do this directly from these definitions with as little extra field-theoretic machinery as possible?
If $E/F$ is a finite field extension, then the field axioms imply that $E$ is a finite-dimensional $F$-vector space. Your definition of basis is exactly that of a vector space, and any two bases of a finite dimensional vector space have the same cardinality.
Proof. Let $A=\lbrace a_1,\dots,a_n\rbrace$ and $B=\lbrace b_1,\dots,b_m\rbrace$. Then $\lbrace a_1,b_1,\dots,b_m\rbrace$ also spans $V$, and we can write $$ b_r=\lambda_1 a_1+\mu_1 b_1+\dots+\mu_{r-1} b_{r-1}+\mu_{r+1} b_{r+1}+\dots+\mu_m b_m, $$ for some $1\le r\le n$ (since $a_1\in A\Rightarrow a_1\ne 0$). Hence $B_1=\lbrace a_1,b_1,\dots,b_{r-1},b_{r+1},\dots,b_m\rbrace$ also spans $V$, and $\# B=\# B_1$.
Now we repeat: $\lbrace a_1,a_2,b_1,\dots,b_{r-1},b_{r+1},\dots,b_m\rbrace$ spans $V$ and we can again remove one of the $b_j$'s and still keep a spanning set.
After doing this $n$ times, we will have $B_n=\lbrace a_1,\dots,a_n,b_{m_1},\dots,b_{m_k}\rbrace$, with $k\ge 0$, $B_n$ spanning $V$, and $\#B = \#B_n$. But $A\subseteq B_n$, and so $\# A\le\# B$. QED.
Now apply to two bases of $E/F$.