Field extensions of finite degree and primitive elements

Let $K\supseteq F$ be fields of characteristic $0$ such that any element of $K$ has degree $\leq n$ over $F$.

If $[K:F]=1$ we are done, and if not then there is some $r_1\in K\setminus F$. Let $E_1=F(r_1)$. Observe that $[E_1:F]\leq n$ by hypothesis. If $[K:E_1]=1$ then $[K:F]=[E_1:F]$ and we are done, and if not then there is some $r_2\in K\setminus E_1$, and let $E_2=E_1(r_2)=F(r_1,r_2)$. Note that because $E_2$ is a finite separable extension of $F$, it is simple, say with $E_2=F(s)$. Thus, again by hypothesis, we have $[E_2:F]\leq n$. It is clear that we can repeat this argument any finite number of times. If we have not found an $m$ for which $K=E_m$ by the time we reach $E_d$ where $d=\lceil\log_2(n)\rceil+1$, then $$[E_d:F]=\underbrace{[E_d:E_{d-1}]}_{\geq 2}\cdots\underbrace{[E_1:F]}_{\geq 2}>n$$ but this contradicts $[E_d: F]\leq n$. Thus $K=E_m$ for some $m$, so that $[K:F]\leq n$.

(I've edited out my previous, unnecessarily-complicated argument using Zorn's lemma. Thanks to Mariano for catching that there was a simpler way.)

As Curtis points out, it is possible for every element of $K$ to have finite degree over $F$, while the extension $K/F$ has infinite degree. For example, take $F=\mathbb{Q}$ and $K=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5},\ldots)$, or $K=\overline{\mathbb{Q}}$. Note that finite is different than bounded.