, 2021

There exists $\alpha \in S$ so that $\{\operatorname{gcd}(\alpha, s) \mid s \in S, s \neq \alpha\}$ contains at least two elements. Since $\alpha$ has only finitely many divisors, there is a $d \mid \alpha$ such that the set $B=\{\beta \in S \mid \operatorname{gcd}(\alpha, \beta)=d\}$ is infinite. Pick $\gamma \in S$ so that $\operatorname{gcd}(\alpha, \gamma) \neq d$. Pick $\beta_{1}, \beta_{2} \in B$ so that $\operatorname{gcd}\left(\beta_{1}, \gamma\right)=\operatorname{gcd}\left(\beta_{2}, \gamma\right)=: d'$. If $d=d'$, then $\operatorname{gcd}\left(\alpha, \beta_{1}\right)=\operatorname{gcd}\left(\gamma, \beta_{1}\right) \neq \operatorname{gcd}(\alpha, \gamma)$. If $d \neq d'$, then either $\operatorname{gcd}\left(\alpha, \beta_{1}\right)=\operatorname{gcd}\left(\alpha, \beta_{2}\right)=d$ and $\operatorname{gcd}\left(\beta_{1}, \beta_{2}\right) \neq d$ or $\operatorname{gcd}\left(\gamma, \beta_{1}\right)=\operatorname{gcd}\left(\gamma, \beta_{2}\right)=d'$ and $\operatorname{gcd}\left(\beta_{1}, \beta_{2}\right) \neq d'$.