If $E$ is generated by $E_1$ and $E_2$, then $[E:F]\leq [E_1:F][E_2:F]$? Suppose $K/F$ is an extension field, and $E_1$ and $E_2$ are two intermediate subfields of finite degree. Let $E$ be the subfields of $K$ generated by $E_1$ and $E_2$. I'm trying to prove that
$$ [E:F]\leq[E_1:F][E_2:F].$$
Since $E_1$ and $E_2$ are finite extensions, I suppose they have bases $\{a_1,\dots,a_n\}$ and $\{b_1,\dots,b_m\}$, respectively. If $E_1=F$ or $E_2=F$, then the inequality is actually equality, so I suppose both are proper extension fields. I think $E=F(a_1,\dots,a_n,b_1,\dots,b_m)$. Since $\{a_1,\dots,a_n,b_1,\dots,b_m\}$ is a spanning set for $E$ over $F$, 
$$[E:F]\leq n+m\leq nm=[E_1:F][E_2:F]$$
since $m,n>1$.
Is this sufficient? I'm weirded out since the problem did not ask to show $[E:F]\leq [E_1:F]+[E_2:F]$ which I feel will generally be a better upper bound.
 A: The space spanned by the union of the two bases is usually not closed under muliplication, so won't span a field.
Instead you should show that
$$
\{a_ib_j\mid 1\le i\le n, 1\le j\le m\}
$$
is a spanning set.
Show that the $F$-space $L$ spanned by that set (inside $K$) is closed under multiplication, and that any field containing both $E_1$ and $E_2$ must contain $L$.
Then show that $L$ also contains the inverse of all its non-zero elements. There the assumption about finiteness of $n$ and $m$ is crucial. Therefore $L$ is the smallest field containing both $E_1$ and $E_2$.
Note that this spanning set is not always linearly independent over $F$. Anyway, the dimension of $L$  is at most $mn$, which is what you wanted to show.
A: The sum of the degrees in general is not going to be an upper bound. Consider $K = \Bbb{Q}(\sqrt[3]{2},e^{2\pi i/3})$. This is a degree $6$ extension of $\Bbb{Q}$. Take $E_1 = \Bbb{Q}(\sqrt[3]{2})$ and $E_2 = \Bbb{Q}(e^{2 \pi i/3})$. Then
$$[E_1 : \Bbb{Q}] + [E_2:\Bbb{Q} ] = 3 + 2 = 5$$
but $E = K$ that has degree $6$ over $\Bbb{Q}$.
