Assume that $F\subset K$ is a quadratic field extension. Choose $\alpha\in K-F$. Then $1,\alpha$ are $F$-linearly independent and since the dimension of $K$ over $F$ is two, they form a basis for $K$ as an $F$-vector space.
I was said that this implies that $K=F(\alpha)$.
1) What does "$=$" mean - equal or isomorphic as $F$-vector spaces?
2) Why is that true (whatever $=$ means)? I do know that if $\alpha$ is algebraic over $F$, then $1, \alpha$ is a basis for $K(\alpha)$ as an $F$-vector space, but I don't know whether $\alpha$ is algebraic over $F$ (if it is, then the spaces $K$ and $F(\alpha)$ are isomorphic because they have the same dimension, but I'm not sure whether they're equal). Update for 2): inspired by this question I can see that $\alpha$ is algebraic over $F$ (because $\alpha^2$ can be expressed a linear combination of $1$ and $\alpha$; in fact it has degree two over $F$), so the $F$-vector spaces $K$ and $F(\alpha)$ are isomorphic by what I said above. But are they equal? If so, why?
This is equality. The extension $F(\alpha)$ is a two-dimensional vector space contained in $K$. Since $K$ is two-dimensional, this must be equality.
Also, the fact that $\alpha$ is algebraic is just because it is an element of a finite extension $K$ of $F$, and all elements of finite extensions are algebraic over the base field.