There are two ways to consider a subfield $K$ of a field $F$ generated by a subset $S \subset F$. The "building up" approach has already been described, so here is the "top-down" approach (I find this more useful in practice).
We can consider all subfields $L \subset F$ such that $S \subset L \subset F$. Then $K = \bigcap_{L \in \mathcal{L}} L$, where $\mathcal{L} = \{ L | L \subset F \text{ is a subfield with } S \subset L \}$. That is, $K$ is the intersection of all the subfields of $F$ which contain $S$.
To see that $K$ is truly a field, we need to notice that it contains all the right things. Certainly $0,1 \in K$ and if $s \in K$, then $s^{-1}$ and $-s$ are in any field containing $s$, so they are in $K$. Similarly, if $s, t \in K$ then $s+t, st \in K$.
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 $p$.)
Best Answer
Let's write $K$ for the field "generated by $1$" and $L$ for the intersection of all subfields of $F$.
It should be clear that $L\subseteq K$. This is because $K$ is a subfield of $F$, and is therefore one of the fields in the intersection that created $L$.
What does it mean to be "generated by $1$", and why is $K\subseteq L$?
Saying that $K$ is generated by $1$ means that $K$ is the smallest subfield containing $1$. Since $K$ is closed under addition, it must contain every element of the form $1 + 1 + \cdots + 1$. Since $K$ is closed under taking negatives, it must contain every element of the form $-(1 + 1 + \cdots + 1)$. There are now two possibilities.
There is some number $n$ such that $$ \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0. $$ In this case, it turns out that $n$ has to be prime and $L=\mathbb F_p$, the finite field with $p$ elements (I'll supply details if you're interested). We don't have to add multiplicative inverses in because they're already present.
There is no number $n$ such that $$ \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0. $$ In this case the subring generated by $1$ is $\mathbb Z$. We think of the number $n$ as being $$ \underbrace{1 + 1 + \cdots + 1}_{n\text{ times}} = 0. $$ You should then check that this definition makes sense with multiplication. (This is a consequence of the distributive law.) But since a subfield must contain multiplicative inverses, we're forced to add these in and we get that $L=\mathbb Q$.
(Note: we say that $F$ has characteristic $p$ in the first case and $0$ in the second.)
Finally, since $L$ is a field it contains $1$ and so it has to also contain the field generated by $1$, as that field is the smallest subfield containing $1$.