Proof: By Euclid
(related to Proposition: 3.10: Two Circles have at most Two Points of Intersection)
 For, if possible, let the circle $ABC$ cut the circle $DEF$ at more than two points, $B$, $G$, $F$, and $H$.
 And $BH$ and $BG$ being joined, let them (then) have been cut in half at points $K$ and $L$ (respectively).
 And $KC$ and $LM$ being drawn at right angles to $BH$ and $BG$ from $K$ and $L$ (respectively) [Prop. 1.11], let them (then) have been drawn through to points $A$ and $E$ (respectively).
 Therefore, since in circle $ABC$ some straight line $AC$ cuts some (other) straight line $BH$ in half, and at right angles, the center of circle $ABC$ is thus on $AC$ [Prop. 3.1 corr.] .
 Again, since in the same circle $ABC$ some straight line $NO$ cuts some (other straight line) $BG$ in half, and at right angles, the center of circle $ABC$ is thus on $NO$ [Prop. 3.1 corr.] .
 And it was also shown (to be) on $AC$.
 And the straight lines $AC$ and $NO$ meet at no other (point) than $P$.
 Thus, point $P$ is the center of circle $ABC$.
 So, similarly, we can show that $P$ is also the center of circle $DEF$.
 Thus, two circles cutting one another, $ABC$ and $DEF$, have the same center $P$.
 The very thing is impossible [Prop. 3.5].
 Thus, a circle does not cut a(nother) circle at more than two points.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"