I read that any field $F$ has a unique smallest subfield $F_0$ (Dummit and Foote : Exercise 7.5.3).
Consider the field $F = F_p \times F_p$. $(F_p,0)$ is a sub-field of $F$ since it has a unit $(1,0)$ and every element has an inverse. Obviously this set is not unique. Other subfields are $\{(0,0), (1,1), ... ,(p,p)\}$ and $(0, F_p)$.
What am I missing here ? Do all subfields of $F$ contain $(1,1)$ ?
The product of two fields is not a field, as it is not even an integral domain (as pointed out in a comment).
But also, yes, a subfield has to contain the multiplicative identity of the original field (this is even the case for subrings, at least this is a common though not universal convention). Thus, what you consider are not even subrings (except for the one containing $(1,1)$). Furthermore this smallest subfield is the field generated by the multiplicative identity. It is either, isomorphic to $Z/pZ$ for a prime $p$ or to the rationals.