$K$ finite extension of $F$ s.t. for every 2 subextensions $M_1, M_2$, $M_1\subset M_2$ or $M_2\subset M_1$. Then there's $a\in K$ such that $K=F(a)$

Let K be a finite extension of a field F such that for every two intermediate field $M_1$, $M_2$ we have $M_1\subset M_2$ or $M_2\subset M_1$. I need to show that there is an element $a\in K$ such that $K=F(a)$.

I have an idea that goes like this: If I show that there are finite intermediate fields, then I could use the Primitive Element Theorem (which demonstration wasn't given during my algebra course, so I would have to include this in my solution so it can be whole).

There's also some doubts here: through the internet we can see that some people enunciate the Primitive Element theorem excluding the hypothesis of the extension (finite) having finite intermediate subfields. Which one is the correct?

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I don't think you need the primitive element theorem. Pick an element $\alpha \in K \setminus F$. If $F(\alpha) = K$ then great; otherwise, choose a $\beta \in K \setminus F(\alpha)$. Using the hypothesis, what's the relationship between $F(\beta)$ and $F(\alpha)$? Try to continue this process. To make the resulting argument formal, you could use induction on the degree of $K$ over $F$.
You probably saw a version of the primitive element theorem that assumes separability, which is a stronger condition: the extension $\mathbf F_p(t^{1/p})$ of the function field $\mathbf F_p(t)$ is inseparable, for example.
The hypothesis implies there's at most one intermediate field of any given degree over $F$ (if there were two, which one would contain the other one?). Can you take it from there?