$f_*$ for a finite morphism preserves perversity, so $f_* f^* E$ is going to be perverse. $f_* f^* E= E \otimes f_( \overline{\mathbb Q_l}$. $f_* (\overline{\mathbb Q_l})$ has the constant sheaf as a summand, (by averaging), so $E$ is a summand of a perverse sheaf, hence is perverse.

Since inseperable morphisms don't change the etale site, we may assume that $f$ is separable, hence finite etale over an open set. By etale descent, over that set $E$ is irreducible. Thus if $E$ is reducible, it has a subobject or quotient object with lower-dimensional support. Call this $F$. It can't be a subobject, because then we would have a nontrivial element in $Hom(F, f_* f^* E)$ giving a nontrivial $Hom(f^* F, f^* E)$ which is impossible because the second is irreducible and their supports are different dimensions. Since $f$ is finite, $f_*= f_!$, and we can apply the same argument to quotient objects: $Hom(f_! f^* F,E)$ is nontrivial, so $Hom(f^* F, f^! E)$ is nontrivial, but again one is irreducible and their supports are different dimensions.