If $A \times B \not = \emptyset$ then there is $a \in A \land b\in B$ so that $\{a,b\} \cap A\times B = \varnothing $ (Axiom of Regularity) I have been stuck showing the following:(this is in the section in the book on the Axiom of regularity)

If $A \times B \not = \emptyset$ then there is $a \in A \land b\in B$ so that $\{a,b\} \cap A\times B = \emptyset $

We see that there is $a\in A$ such that $\{a\} \cap A\times B = \emptyset$ Since if that does not hold then we see that $A\subseteq A\times B$ which would mean $A = \emptyset$ (already proven) which can't happen Since $A\times B \not=\emptyset$
Thus to finish the proof:
we need to show that there is a $b \in B$ such that $\{b\} \cap A\times B = \emptyset$
but i can't seem to do this… is it true that if $B \subseteq A \times B$ then $B = \emptyset$?
 A: You have defined $(x, y) = \{x, \{x, y\}\}$ in the comments.
Consider $C = \{\{a, b\} \mid a \in A, b \in B\} \cup (A \times B)$. By the axiom of regularity, take some $c \in C$ such that $c \cap C = \emptyset$.
Since $c \in C$, we have two possibilities. The first is that $c$ is of the form $\{a, b\}$ for some $a \in A$ and $b \in B$. In this case, we have $\{a, b\} \cap (A \times B) \subseteq \{a, b\} \cap C = \emptyset$, and we are done. The second possibility is that $c \in A \times B$. Then write $c = (a, b)$ for some $a \in A$, $b \in B$. Then $\{a, b\} \in c$ and $\{a, b\} \in C$, so $\{a, b\} \in c \cap C$; this contradicts the fact that $c \cap C = \emptyset$.
A: Here is a proof of your desired claim. We proceed by contradiction, so assume your desired claim is false. That is, assume that $A\times B\not=\emptyset$ and
$$(1)\ \forall a\in A \forall b\in B ( \{a,b\}\cap A\times B \not=\emptyset)$$
But $(1)$ is equivalent to
$$(1')\ \forall a\in A \forall b\in B ( a\in A\times B\lor b\in A\times B)$$
As you have already noted (and I will assume without proof), there is an $a_0\in A$ such that $a_0\not\in A\times B$. Instanting $(1')$ on $a_0$ lends:
$$(2)\ \forall b\in B (a_0\in A\times B  \lor b\in A\times B)$$
But since $a_0\not\in A\times B$, we derive:
$$(2')\ \forall b\in B (b\in A\times B)$$
Now we seek to show that $(2')$ contradicts the axiom of regularity. I will do this by showing that a more general situation contradicts the axiom of regularity, and then showing that what we have is an instance of the more general situation. The general situation is the following:
Let $X$ be a non-empty set with the following property:
$$(3)\ \forall x\in X\exists c_x\exists y_x(y_x\in X\land (y_x\in c_x\in x))$$
We can use this property to define an infinite descending chain of sets. An infinite descending chain of sets is a function $f : \mathbb{N}\to V$ (where $V$ is the universe, or some other suitable set) with the property that $f(n+1)\in f(n)$, for each $n\in \mathbb{N}$. The function I will define will also have the property that, for $n=2k$, $f(n)\in X$ (this will be important for the inductive step of the definition).
Since $X$ is non-empty, we have some $x_0\in X$. So we may define $f(0):=x_0$. Now, since $x_0\in X$, we may instantiate $(3)$ on $x_0$ to obtain:
$$(3')\ \exists c_{x_0}\exists y_{x_0}(y_{x_0}\in X\land (y_{x_0}\in c_{x_0}\in x_0))$$
Now, we use choice to select some such $c_{x_0}$ and $y_{x_0}$. We set $f(1):= c_{x_0}$ and $f(2):=y_{x_0}$. Now for the inductive step. We will handle the case that $n$ is even and that $n$ is odd separately. For $n=2k$, we have that $f(n)\in X$. So we instantiate $(3)$ on $f(n)$ to obtain:
$$(3'')\ \exists c_{f(n)}\exists y_{f(n)}(y_{f(n)}\in X\land (y_{f(n)}\in c_{f(n)}\in f(n))$$
We again use choice and set $f(n+1):=c_{f(n)}$ and $f(n+2):=y_{f(n)}$. Now we handle the case when $n=2k+1$. In this case, we have that $n+1=2(k+1)$, so we know $f(n+1)\in X$. So we instantiate $(3)$ on $f(n+1)$ to obtain:
$$(3'')\ \exists c_{f(n+1)}\exists y_{f(n+1)}(y_{f(n+1)}\in X\land (y_{f(n+1)}\in c_{f(n+1)}\in f(n+1)))$$
We again use choice so that we can set $f(n+2):=c_{f(n+1)}$ and $f(n+3):=y_{f(n+1)}$. This completes the definition of $f$. The way I defined $f$ is not technically proper, but it is hopefully clear that we could indeed define $f$ in a proper manner. It is not too hard to verify that $f(n+1)\in f(n)$, for all $n\in\mathbb{N}$. This contradicts the axiom of regularity. Thus, no such set $X$ can exist.
All that remains to show is that our problem is an instance of this situation. Now, from $(2')$ and the definition of $A\times B$ (using the Kuratowski definition), we obtain:
$$(2'')\ \forall b \in B\exists a'\in A\exists b'\in B (b=\{\{a'\},\{a',b'\}\})$$
We take $c_b:=\{a',b'\}$ and $y_b:=b'$. This indeed lends
$$\forall b\in B\exists  c_b\exists y_b(y_b\in B\land (y_b\in c_b\in b))$$
This completes the proof.
