Three circles intersect pairwise but let's assume there is no point shared by all three of them. There are three pairs of circles. Two circles in any pair share a chord. The problem is to prove that the three chords meet at a point.
This page gives a solution:
Consider each circle as the equator of a sphere. Let the plane of the circles be horizontal. It cuts the spheres in two (of course symmetric) halves so that a circle is the projection of the corresponding sphere onto the horizontal plane. Any two spheres intersect at a circle. The chords at hand are the projections of three such (vertical) circles. However, the three spheres share two points which are symmetric with respect to the plane. Those points belong to all three pairwise intersections of the spheres - vertical circles. Therefore, projections of the three vertical circles - our chords - share a point which is the projection of the common points of the spheres.
The part(in bold type) I don’t understand is why it is certain that the three vertical circles(pairwise intersection of the spheres) intersect at a common vertical line. Isn’t it possible that these three vertical circles only intersect pairwise but don’t have common intersection by all three?
Since the sphere argument seems a bit involute, I give you a very simple proof based on radical axis. Recall that the radical axis of two circle is the locus of points which have the same power respect to both circles.
Let $\Gamma_1,\Gamma_2,\Gamma_3$ be the three given circles, $r_{ij}$ the radical axis of $\Gamma_i$ and $\Gamma_j$ for each $i<j$. Let $P\in r_{12}\cap r_{23}$ and let $p_i$ denote the power of $P$ respect to $\Gamma_i$. Then $P\in r_{12}$ implies $p_1=p_2$ while $P\in r_{23}$ implies $p_2=p_3$, thus by transitivity, $p_1=p_3$, hence $P\in r_{13}$. This proves $P\in r_{12}\cap r_{23}\cap r_{13}$.