The typical example of a polynomial that is not separable: let $F=\mathbb{F}_p(x)$ be the field of all rational functions with coefficients in the field with $p$ elements. Consider the polynomial $y^p-x$ in $F[y]$. If $u$ is a root of $y^p-x$ (in some extension of $F$), we have $u^p-x=0$, so $u^p=x$. Because we are in charactersitic $p$, the binomial theorem becomes
$$(a\pm b)^p = a^p \pm b^p,$$
so therefore we have
$$(y-u)^p = y^p-u^p = y^p-x,$$
so $u$ is the only root of $y^p-x$ (by unique factorization), which is a perfect $p$ power in its splitting field.
Note also that $y^p-x$ is irreducible over $F$ (so that $u$ actually "lives" in some extension strictly larger than $F$). To prove this, consider e.g. by Eisenstein's criterion applied to it as an element of $(\mathbb{F}_p[x])[y]$: $\mathbb{F}_p[x]$ is a UFD where $x$ is prime; every coefficient of $y^p-x$ except the leading one is a multiple of $x$, and the constant term is not divisible by $x^2$; so by Eisenstein's Criterion $y^p-x$ is irreducible in $(\mathbb{F}_p[x])[y]$. Now Gauss's Lemma lets us conclude that the polynomial is also irreducible over $F$.
Over a field of characteristic $0$, all irreducible polynomials are separable. This follows because a polynomial has a multiple root if and only if the gcd with its formal derivative is not equal to $1$. But if $p(x)$ is irreducible, then $p'(x)$ is a polynomial of strictly smaller degree than $p(x)$, and therefore no nonconstant factor of $p(x)$ can be a factor of $p'(x)$. Thus, $\gcd(p(x),p'(x))=1$, so $p(x)$ is separable.
So inseparability is a purely positive characteristic phenomenon. In fact, one can show that an irreducible polynomial over a field of characteristic $p$ is not separable if and only if it can be written as a polynomial in $x^p$ (that is, every exponent that occurs is a multiple of $p$).
