Use the C rule and proposition 2.10 to prove $$\forall x(A(x)\to B(x)\lor C(x))\land\lnot\forall x(A(x)\to B(x)))\vdash\exists x(A(x)\land C(x))$$
[ C rule ]
[ A4 rule ]
This is what I've done so far
By hypothesis,$\forall x(A(x)\to B(x)\lor C(x))\land\lnot\forall x(A(x)\to B(x)))$, particularly $$\lnot\forall x(A(x)\to B(x)))$$ $$\implies\exists\lnot x(A(x)\to B(x)))$$
Also by hypothesis, $$\forall x(\lnot A(x)\lor B(x)\lor C(x))$$
How do I proceed from here? Should I apply C rule or A4 rule?
I don't know how to get $\exists x(A(x)\land C(x))$
By C-rule, exists a new constant $c$ such that $\vdash_C\neg(A(c)\rightarrow B(c))$ and this is equivalent to $\vdash_CA(c)\wedge \neg B(c)$, so $\vdash_CA(c)$.
As hypothesis you have that $\vdash\forall x((A(x)\rightarrow B(x))\vee C(x))$. Then you can conclude that $\vdash_CC(c)$ (here I use A.4 rule on the previous statement).
Now, by $\exists$-introduction you have $\vdash_C\exists x(A(x)\wedge C(x))$. The conclution follows by proposition 2.10.