A subgroup $A\times B$ of $\mathbb{Z}\times\mathbb{Z}$ with the property that either $A\not <\mathbb{Z}$ or $B\not <\mathbb{Z}$. I think I have found a subgroup $A\times B$ of $\mathbb{Z}\times\mathbb{Z}$ with the property that either $A\not <\mathbb{Z}$ or  $B\not <\mathbb{Z}$. I wanted to see if my counterexample is a valid counterexample.
Let $$K=\{(x,-x):x\in \mathbb{Z}\}.$$ Then $(0,0)\in K$, and if $(x,-x),(y,-y)\in K$, then, by definition of $K$, $(-y,y)\in K$ and therefore $(x-y,(-x+y))=(x-y,-(x-y))\in K$.
For any group $G$, $H<G$ if and only if $xy^{-1}\in H$ whenever $x,y\in H$ and so $K<\mathbb{Z}\times\mathbb{Z}$.
Now, write $K=A\times B$ and observe that if $A<\mathbb{Z}$ and $B<\mathbb{Z}$, then, since $1\in A$ and $1\in B$, we have that $A=\mathbb{Z}$ and $B=\mathbb{Z}$. But then $A\times B=\mathbb{Z}\times \mathbb{Z}$ and, in particular, $(1,2)\not\in K=A\times B$. $\blacksquare$
Ultimately I am trying to show that it is not the case that $A <\mathbb{Z}$ and $B <\mathbb{Z}$ iff $A\times B<\mathbb{Z}\times\mathbb{Z}$.
 A: Even if you completely ignore the group structure, there are no sets $A$ and $B$ so that $K=A\times B$.
Proof:
Assume $K=A\times B$. Then, by definition of the cross product, $k\in K$ iff there exist $a\in A$ and $b\in B$ such that $k=(a,b)$.
Now $(-1,1)\in K$, therefore $-1\in A$ and $1\in B$. Also, $(1,-1)\in K$ therefore $1\in A$ and $-1$ in $B$.
But if $1\in A$ and $1\in B$, then $(1,1)\in A\times B$. But $1\ne -1$, therefore $(1,1)\notin K$, in contradiction to the assumption $K=A\times B$. $\square$
So while you correctly proved that that $K$ is a subgroup of $\mathbb Z\times\mathbb Z$, it is not of the form $A\times B$, therefore it is not a valid counterexample.

Another way to interpret your question is whether $K$ is isomorphic to a group $A\times B$ with $A<\mathbb Z$ and $B<\mathbb Z$. In that case, the answer is: It is isomorphic to the group $\mathbb Z\times\{0\}$ via the isomorphism $(x,-x)\mapsto (x,0)$. Of course both $\mathbb Z$ and $\{0\}$ are subgroups of $\mathbb Z$. Therefore also in this interpretation, $K$ is not a counterexample.
