As mentionned, free groups have many non-finitely-generated subgroups. Commutator subgroups are such an example (see here), but more generally it is known that, in a free group of finite rank, a normal subgroup is finitely-generated if and only if it is a finite-index subgroup (see here). For instance, we know that the normal closure $\langle \langle a \rangle \rangle$ is a non-finitely-generated subgroup of the free group $\mathbb{F}_2= \langle a,b \mid \ \rangle$.
In fact, the commutator subgroup of any free product $A \ast B$ turns out to be free of infinite rank whenever $A$ or $B$ is infinite (see here).
More generally, any countable group is embeddable into a 2-generator group (see here for a proof). Therefore, any countable non-finitely-generated group (such that $\mathbb{Q}$, the free group $\mathbb{F}_{\infty}$ of infinite rank or the group $S_{\infty}$ of bijections $\mathbb{N} \to \mathbb{N}$ with finite support) appears as an infinite-index subgroup of a finitely-generated group.
Another source of examples comes from wreath products. For instance, let $L_2$ denote the lamplighter group. Then $$L_2 = \left( \bigoplus\limits_{n \in \mathbb{Z}} \mathbb{Z}_2 \right) \rtimes \mathbb{Z},$$ so the subgroup $\bigoplus\limits_{n \in \mathbb{Z}} \mathbb{Z}_2$ is clearly not finitely-generated whereas $L_2$ turns out to be finitely-generated.
Separability is automatic in characteristic 0 and finite fields.
Let $f(x) \in \mathbb{F}[x]$. Let $f'(x)$ be its formal derivative. Then $f(x)$ has a repeated root (in some extension field) iff $f(x)$ and $f'(x)$ fail to be relatively prime.
If you take a field of characteristic $0$, non-constant polynomials have nonzero derivatives (of degree one less). Thus if $f(x)$ is irreducible, then $f'(x)$ must be relatively prime to $f'(x)$: If $g(x)$ divides $f(x)$ it's either $1$ or $f(x)$ up to associates. But $f'(x)$ cannot be divisible by $g(x)=f(x)$ -- it's degree is too small. Thus any common divisor $g(x)$ must be $1$.
This means that irreducible polynomials in fields of characteristic 0 cannot have repeated roots. Thus in characteristic 0 everything is separable (this explains why it's hard to think up inseparable stuff off the top of your head -- we tend to work in char 0 most of the time).
As for finite fields, every element in $\mathbb{F}_q$ (the field of order $q=p^n$ some prime $p$) is a root of $f(x) = x^q-x$. Thus the minimal polynomial of any element of a finite field must be a divisor of $f(x)$. Notice that $f'(x)=qx^{q-1}-1=-1$ (since $q=0$ in char $p$). The gcd of $f(x)$ and $f'(x)=-1$ is $1$ so $f(x)$ has no repeated roots. This means its factors cannot have repeated roots either. Thus every element in a finite field is separable.
Therefore, if you're looking for something that isn't separable, you'll need an infinite field of characteristic $p \not=0$. The canonical example is...
Consider $\mathbb{F}=\mathbb{Z}_p(y)$ (rational polynomials in $y$ with coefficients in $\mathbb{Z}_p$). Let $f(x) = x^p-y \in \mathbb{F}[x]$. [Notice that $f'(x)=px^{p-1}=0$.]
Now $f(x)$ is irreducible by Eisenstein's criterion: $y$ is prime in $\mathbb{F}$ since $\mathbb{F}/(y) \cong \mathbb{Z}_p$ (an integral domain) so $(y)$ is a prime ideal and so $y$ (being a nonzero generator of a prime ideal) is prime. Notice that $y$ divides all but the leading coefficients of $f(x)=x^p-y$ and $y^2$ does not divide $f(0)=y$ (the constant term).
Next, adjoin a root of $f(x)$. Let's call this root $\alpha$ (or in more suggestive notation we could say $\alpha=\sqrt[p]{y}$). This means that $f(\alpha)=\alpha^p-y=0$ so $\alpha^p=y$. Notice that $(x-\alpha)^p = x^p - \alpha^p$ (the middle binomial coefficients are divisible by $p$ and so are $=0$). Therefore, $f(x)=x^p-y = (x-\alpha)^p$. We have an irreducible polynomial with a single ($p$-times) repeated root. $y$ is inseparable!
Best Answer
If $K/F$ is finite, say $n = [K:F] < \infty$, then (by definition) $K = \text{Span}_F\{k_1,\ldots,k_n\}$ where $k_i \in K$ for each $i$. Hence every element of $K$ can be written as a finite linear combination of the $k_i$ over $F$, and so we know $K \subseteq F[k_1,\ldots,k_n]$. At the same time $F[k_1,\ldots,k_n] \subseteq K$ is clear since $K$ is closed under addition and multiplication. So $K = F[k_1,\ldots,k_n] = F(k_1,\ldots,k_n)$ is finitely generated.