Proving a set theoretic axiom counter positively. I'm given 2 axioms from set theory:
${\cal F}$: For each set $A\neq \emptyset$ $\exists$ a set $B$ such that $B\in A$ and $B\cap A = \emptyset$
${\cal S}$: $\nexists$ a sequence $(x_n)_{n\in \mathbb{N}}$ such that $\forall n \in \mathbb{N}$ we have $x_{n+1}\in x_n$.
I'm to show that ${\cal S} \to {\cal F}$ by arguing counter positively.
I'm more than a little stuck, but am I supposed to begin like this:
We assume
$\neg {\cal F}$: $\exists$ a set $A\neq \emptyset$ such that $\forall$ sets $B\in A$ $A\cap B \neq \emptyset$
Next we assume that ${\cal S}$ is true. And this is where I get stuck.

Fix $A \neq \emptyset$ such that there is no $B \in A$ with $A \cap B = \emptyset$.
Fix any $x_1 \in A$. By our assumption we have that $A \cap x_1 \neq \emptyset$. Hence we may fix some $x_2 \in A \cap x_1$. But, again by our assumption on $A$, we must have $A \cap x_2 \neq \emptyset$ and thus there is some $x_3 \in A \cap x_2$.
Now show that this procedure (using the axiom of dependent choice) allows us to construct a sequence $(x_n \mid n \in \mathbb N)$ such that $\forall n \in \mathbb N \colon x_{n+1} \in x_n$.
