Irreducible polynomials have distinct roots?

Consider a field $F$ of characteristic $p$. A polynomial has multiple roots only if it has a root in common with its (formal) derivative; that is, the multiple roots of $f$ are the roots of $\gcd(f,f')$. Since $f$ is irreducible, multiple roots can occur only if the $\gcd$ is $f$ itself, that is $f'$ is a multiple of $f$. And that is only possible if $f'=0$, that is, all monomials in $f$ have degree a multiple of $p$, so $f(x)=g(x^p)$ for some polynomial $g$.

If $F$ is finite, then $\phi\colon a\mapsto a^p$ is an automorphism of $F$ (and also of the splitting field $E$ of our polynomial), and there exists $h(x)$ such that $\phi(h)=g$. Then for $\alpha\in E$ with $f(\alpha)=0$ also $h(\alpha)=0$ (because $\phi(h(\alpha))=\phi(h)(\phi(\alpha))=g(\alpha^p)=f(\alpha)=0$). Since $h$ is of smaller degree than $f$, $f$ is not irreducible.

As this proof shows, one has to look for cases where $\phi$ is not an automorphism to find a counterexample (such as in Andreas Carantis comment).

The result holds over any finite field. One way of seeing this is that if $h(x) \in \Bbb{F}_{q}[x]$ (where $q$ is a power of the prime $p$) is irreducible over $\Bbb{F}_{q}$, and $\alpha$ is one of its roots, then $\alpha$ is also algebraic over $\Bbb{F}_{p}$. If $f(x) \in \Bbb{F}_{p}[x]$ is the minimal polynomial of $\alpha$ over $\Bbb{F}_{p}$, then $h(x)$ divides $f(x)$, and you know that the latter has distinct roots.

There are examples, though, of irreducible polynomials of degree $> 1$ over an infinite field of positive characteristic which have only one root.

As pointed out, to get an example of an irreducible polynomial with multiple roots, it cannot be finite. Here is an example where this happens in an infinite field of characteristic $p$.

Consider the irreducible polynomial $f(x)=x^p-t\in \mathbb{F}_p(t)[x]$ in the extension field $\mathbb{F}_p(t^{1/p})$ of the field $\mathbb{F}_p(t)$ of rational functions over $\mathbb{F}_p$.

Notice that $(x - t^{1/p})^p = x^p + (-1)^p t = x^p - t$, where the first equality follows from the binomial theorem and the second is obviously true for all odd $p$, and when $p=2$ we have $t=-t$ so it holds there as well.

Thus $f$ has one root of multiplicity $p$ in $\mathbb{F}_p(t^{1/p})$.