You could go the distance

$a+b+c=0$ ..... $(1)$
$a^2+b^2+c^2=4$ ..... $(2)$

Squring both sides to $(1)$ .

$\Rightarrow a^2+b^2+c^2+2(ab+bc+ca)=0$
$\Rightarrow (ab+bc+ca)=-2$

Now, squring both sides to $(2)$ .
$\Rightarrow a^4+b^4+c^4+2[(ab)^2+(bc)^2+(ca)^2]=16$
$\Rightarrow a^4+b^4+c^4+2[(ab+bc+ca)^2]=16$
$\Rightarrow a^4+b^4+c^4+8=16$
$\Rightarrow a^4+b^4+c^4=\boxed{8}$

---

Note: $(ab)^2+(bc)^2+(ca)^2=(ab+bc+ca)^2$
$\Rightarrow (ab)^2+(bc)^2+(ca)^2=(ab+bc+ca)^2-2abc(a+b+c)$
From $(1),$ $(a+b+c)=0$ .
$\Rightarrow (ab)^2+(bc)^2+(ca)^2=(ab+bc+ca)^2$