Let me address your first question. First, I want to argue that there is no precise meaning of "involving numbers only". For example, given a finite field $F$ of size $4$ constructed in the usual manner (quotient of a polynomial ring over $\mathbb{Z}/2\mathbb{Z}$), I can choose a set of numbers, say
$$S=\{37,\tfrac{5}{19},\pi,e\}$$
and, choosing a bijection of $S$ with $F$, use transport of structure to give $S$ the structure of a field. The field structure does not depend in any way on what the underlying set is "made of".
However, along the lines of what I think you are ultimately after, you can obtain finite fields of any possible order using larger rings of integers. For example, $\mathbb{Z}[i]/(3)$ is a finite field of size $9$, and $\mathbb{Z}[i]$ consists of very reasonable numbers,
$$\mathbb{Z}[i]=\{a+bi\mid a,b\in\mathbb{Z}\}.$$
Now let me addressr your second question. Let's use $\mathbb{F}_p$ to mean $\mathbb{Z}/p\mathbb{Z}$, a finite field of order $p$ - it is a very common notation that is slightly less cumbersome, but doesn't mean anything different, they are exact synonyms.
A finite field of order $p^n$ is often constructed by taking the polynomial ring $\mathbb{F}_p[x]$, choosing an irreducible polynomial $f\in \mathbb{F}_p[x]$ of degree $n$, and then making the field
$$F=\mathbb{F}_p[x]/(f).$$
Now, the division algorithm for polynomials tells you that each equivalence class in this quotient can be uniquely identified by a representative of degree $<n$. In other words,
$$\begin{align*}
F&=\{a_0+a_1x+\cdots +a_{n-1}x^{n-1}+(f)\mid a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\}\\\\
&=\left\{\,\overline{a_0+a_1x+\cdots +a_{n-1}x^{n-1}}\,\;\middle\vert\;a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\right\}\\\\\\
&=\{a_0+a_1\overline{x}+\cdots +a_{n-1}\overline{x}^{n-1}\mid a_0,a_1,\ldots,a_{n-1}\in\mathbb{F}_p\}
\end{align*}$$
Letting the symbol $\alpha$ be a stand-in for $\overline{x}$, you can think of $F$ as being $\mathbb{F}_p$ with a new element "$\alpha$" added in, where $\alpha$ is a root of $f$, and you can write $F=\mathbb{F}_p[\alpha]$.
Now, the prime subfield of $F$ is just the "constant" polynomials, i.e. the ones with no $\alpha$'s in them:
$$\text{the prime subfield of }F=\{a_0+0\alpha+\cdots+0\alpha^{n-1}\mid a_0\in\mathbb{F}_p\}$$
and for each divisor $d\mid n$, the unique subfield of $F$ of order $p^d$ is the collection of polynomials in $\alpha$ whose terms are those of exponents that are multiples of $n/d$:
$$\text{the subfield of }F\text{ of order }p^d=\{a_0+a_1\alpha^{n/d}+\cdots+a_{d-1}\alpha^{(d-1)n/d}\mid a_0,a_1,\ldots,a_{d-1}\in\mathbb{F}_p\}$$
(clearly, the above set has cardinality $p^d$, because it takes $d$ elements of $\mathbb{F}_p$ to specify a given element of the above set, namely, each of the coefficients of the powers of $\alpha$. To see that it is a field, remember that $(a+b)^p=a^p+b^p$ in a field of characteristic $p$.)
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 $\Bbb Q$.
Best Answer
Assume char(F)$=p$ to start with and let $e$ be the multiplicative unit in $F$. Let $A$ be a subfield of $F$. According to subfield axioms, $A$ contains $0$, $e$, $2e$, ..., $(p-1)e$ which are distinct elements, hence $$\{0,e,...,(p-1)e\}\subset A$$ for every subfield $A$. The set on the left is itself a subfield of $F$ so it must be the intersection of all subfields. It should be clear that it is isomorphic to $\mathbb{F}_p$.
The case where the characteristic is infinite is similar except you can show that every subfield must contain $ne$ where $n\in\mathbb{N}$ and hence must contain $re$ where $r\in\mathbb{Q}$.