4. As shown in Figure 2, in Rt $\triangle A B C$, $A F$ is the altitude, $\angle B A C=90^{\circ}$, and $B D=D C=F C=$ 1. Then $A C$ is ( ).
(A) $\sqrt[3]{2}$
(B) $\sqrt{3}$
(C) $\sqrt{2}$
(D) $\sqrt[3]{3}$
4. (A).

Let $A C=x, B D=C D=C F=1, A D=x-1$. By the Pythagorean theorem, we have
$$
A B^{2}=B D^{2}-A D^{2}=1^{2}-(x-1)^{2} \text {. }
$$

By the projection theorem, we get
$$
B C=\frac{A C^{2}}{C F}=x^{2} \text {. }
$$

By the Pythagorean theorem, we have
$$
A B^{2}+A C^{2}=B C^{2} \text {, }
$$

which means $1^{2}-(x-1)^{2}+x^{2}=x^{4}$, or $x^{4}=2 x$.
Since $x>0$, then $x^{3}=2$, hence $x=\sqrt[3]{2}$.