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!
If you want to avoid the primitive element theorem : For each root $\beta$ of $f \in F[x]$ your separable polynomial whose $E$ is the splitting field, if $\beta \not \in F$ then it has a distinct $F$-conjugate $\gamma$, let $\sigma : F(\beta) \to F(\gamma)$ be the natural field homomorphism, it can be extended to an homomorphism $\sigma:E \to \sigma(E) \subset \overline{E}$, since $E/F$ is normal then $\sigma(E) = E$ and hence $\sigma\in Gal(E/F)$ and $\beta \not \in E^{Gal(E/F)}$.
That is to say $E^{Gal(E/F)}=F$ and $E/F$ is Galois.
Best Answer
If you just multiply all the minimal polynomials then it might not be separable, because maybe two elements $\alpha_i,\alpha_j$ have the same minimal polynomial. But you can take the distinct minimal polynomials of the elements $\alpha_1,...,\alpha_k$, say $f_1, f_2,...,f_s$ and then take the splitting field of $f=f_1f_2...f_s$. This splitting field is the same as the splitting field of the polynomial you have defined, the only difference is that this polynomial is indeed separable. (which proves the extension is Galois). It is separable because of the following two facts:
$1$. All the polynomials $f_1,...,f_s$ are separable because $K/F$ is a separable extension.
$2.$ If $i\ne j$ then $f_i$ and $f_j$ have no common roots in the algebraic closure. This is because two distinct monic irreducible polynomials are coprime. And it is a standard result that if two polynomials are coprime over $F$ then they are coprime over any extension of $F$, including the algebraic closure.
Combine these two facts, it clearly follows that $f$ can't have a multiple root.