Recall the quartic version of the Cayley-Bacharach theorem:

**Theorem:** Consider two quartics in general position, which intersect in $16$ points (by Bezout's theorem). Then if a third quartic passes through $13$ of those points, it also passes through the other $3$.

This can be used to establish a simpler result:

**Lemma:** Let $\Gamma$ be a cubic, and let $T_1, T_2, T_3, T_4$ be triples of points on $\Gamma$. Suppose that for each $i \in \{1, 2, 3\}$, the six points $T_i \sqcup T_{i+1}$ lie on a conic $C_i$. Then the six points $T_1 \sqcup T_4$ also lie on a conic.

**Proof:** We have an obvious set of $12$ points, and wish to extend it to a set of $16$ points so that Cayley-Bacharach is applicable. We let $P$ be the fourth intersection point of $C_1$ and $C_2$ (i.e. the one that isn't in $T_2$), and $Q$ be the fourth intersection point of $C_2$ and $C_3$. Now draw the line $l$ through $P$ and $Q$; it intersects $C_1$ again at a point $R$, and $C_3$ again at a point $S$.

Then $T_1 \sqcup T_2 \sqcup T_3 \sqcup T_4 \sqcup \{P, Q, R, S \}$ is a set of $16$ points. There are two quartics passing through it, namely $C_1 \cup C_3$ and $\Gamma \cup \ell$. Note that $C_2$ passes through $8$ of these $16$ points; let $C_4$ be the conic passing through any $5$ of the other $8$ points. Then $C_2 \cup C_4$ is a quartic passing through $13$ of the points, so passes through the other $3$ by Cayley-Bacharach. The result follows.

**Corollary:** Let $\Gamma$ be a cubic, and let $T_1, T_2, \dots, T_{2k}$ be triples of points on $\Gamma$. Suppose that for each $i \in \{1, 2, \dots, 2k - 1\}$, the six points $T_i \sqcup T_{i+1}$ lie on a conic $C_i$. Then the six points $T_1 \sqcup T_{2k}$ also lie on a conic.

**Proof:** Repeated application of previous result.

Your theorem is the case $k = 3$.