Maybe an elementary proof that I gave in a more general context, but I can't find it on the site, so I'll adapt it to this case.
Set $n=\lfloor \sqrt p\rfloor$. Suppose $\sqrt p$ is rational and let $m$ be the smallest positive integer such that $m\sqrt p$ is an integer. Consider $m'=m(\sqrt p-n)$; it is an integer, and
$$ m'\sqrt p=m(\sqrt p-n)\sqrt p=mp-nm\sqrt p $$
is an integer too.
However, since $0\le \sqrt p-n <1$, we have $0\le m' <m $. Since $m$ is the smallest positive integer such that $ m\sqrt p$ is an integer, it implies $m'=0$, which means $\sqrt p=n$, hence $p=n^2$, which contradicts $p$ being prime.