A field with characteristic $0$ contains $\mathbb Q$

If you think the subfield test is wrong then you should go back and review why it's sound. Indeed, the test tells us $0=a-a\in A$, hence $0-a\in A$ for all $a\in A$, so it's closed under additive inverses, and $1=aa^{-1}\in A$ for any nonzero $a\in A$, hence $1\in A$ and $1a^{-1}\in A$ for all nonzero $a\in A$, so it's also closed under multiplicative inverses. Addition is $a-(0-b)$ and multiplication $a(1b^{-1})^{-1}$, so the subset is closed under addition, subtraction, multiplication, division, it has $0$ and $1$, so it is a subfield. Hence we do know $ab^{-1}+cd^{-1}\in A$ if $a,b,c,d\in A$.

You say you don't think $T$ should be a subfield, but your intuition should suggest the opposite conclusion: $T$ is the set of all ratios between sums of $1$ and their additive inverses, which is exactly what the rationals $\Bbb Q$ are! This doesn't require the subfield test per se: one may argue that $T$ is a field in the exact same way that one argues $\Bbb Q$ is a field! Should be a trip down memory lane.

As for the intersection discussion: the prime subfield of a field is the minimal field, which has no smaller subfields, and one may prove that it is equal to the intersection of all subfields. Since $1$ is contained in all subfields, so are its sums and additive inverses, and their ratios, all by the closure property of the operations, so it follows that $T$ is contained in all subfields hence their intersection, hence the intersection equals $T$, hence $T$ is a prime subfield. Since $T\cong\Bbb Q$, this means we have shown that the field contains a prime subfield which is just a copy of $\Bbb Q$.

Well, that subring $S$ is not only isomorphic to $\Bbb Z$, but it is generated by the unit element $1$ of the field $F$. More concretely $S$ consists of $\pm(1+1+\dots+1)$ which are operations in $F$.

Yes, we have to verify that the given map also respects addition. But, using commutativity of multiplication we can imitate the common denominator: $$ab^{-1}\,+\,cd^{-1}=adb^{-1}d^{-1}+bcb^{-1}d^{-1}=(ad+bc)b^{-1}d^{-1}\,.$$

I think the cited paragraph wants to say that $F$ not only contains $\Bbb Q$, but it is its smallest subfield, i.e. the intersection of all of its subfields.