Euclidean Version of Pappus's theorem

I'm not sure what Hartshorne has in mind, but Pappus' theorem is a simple consequence of similarity of Euclidean triangles (in guise of the intercept theorem) and there's no need of introducing the circle:

Let $P$ be the point of intersection of $l$ and $m$. All we need to do is to show that $\frac{PA'}{PB} = \frac{PB'}{PA}$.

But by assumption we have $\frac{PA'}{PC} = \frac{PC'}{PA}$ and $\frac{PC}{PB} = \frac{PB'}{PC'}$. Multiplying the left hand sides and the right hand sides together, we get what we want.

Without looking in too much detail, I suspect that you have not yet used the property of cyclic quadrilaterals that opposite angles are supplementary. For example, considering the cyclic quadrilateral $ADB'C'$, $\angle B'AD$ and $\angle B'C'D$ are supplementary, so $\angle B'AD\cong\angle DC'A'$.

As a further suggestion, though I don't immediately see how to prove it, the circle through $A$, $B'$, and $C'$, the circle through $A'$, $B$, and $C'$, and the circle through $A'$, $B'$, and $C$ all pass through that same point $D$.