In the following, we call a three-element subset $\{s, t, u\} \subset S$ a balanced triangle if the set $\{\mathrm{ggT}(s, t), \operatorname{ggT}(s, u), \operatorname{ggT}(t, u)\}$ has exactly two different elements. It is to be shown that there exists a balanced triangle.

Lemma. For pairwise different numbers $a, b, c, d \in S$, such that $\operatorname{ggT}(a, b)=\operatorname{ggT}(a, c) \neq \operatorname{ggT}(a, d)$ and $\operatorname{ggT}(b, d)=\operatorname{ggT}(c, d)$, the set $\{a, b, c, d\}$ contains a balanced triangle.
Proof. If $\operatorname{ggT}(a, b)=\operatorname{ggT}(b, d)$, then $\{a, b, d\}$ is a balanced triangle. Otherwise, either $\operatorname{ggT}(a, d) \neq \operatorname{ggT}(a, b)$ or $\operatorname{ggT}(a, d) \neq \operatorname{ggT}(b, d)$, so $\{a, b, c\}$ or $\{b, c, d\}$ is a balanced triangle.
!

For each element $a \in S$, let $S_{a}=\{\operatorname{ggT}(a, s) \mid s \in S, s \neq a\}$. Since this set contains only divisors of $a$, it is finite. From the assumption, we can choose $a \in S$ such that $S_{a}$ has at least two elements, otherwise $\operatorname{ggT}(v, w)=\operatorname{ggT}(w, x)=\operatorname{ggT}(x, y)$ would hold.

By the pigeonhole principle, we find an infinite subset $T \subset S$ such that $\operatorname{ggT}(a, t)$ is the same value $g$ for all $t \in T$. Now we choose a $d \in S \backslash(T \cup\{a\})$ such that $\operatorname{ggT}(a, d) \neq g$, which must exist due to $\left|S_{a}\right|>1$. Since $S_{d}$ is also finite, we find by the pigeonhole principle two different elements $b, c \in T$ such that $\operatorname{ggT}(b, d)=\operatorname{ggT}(c, d)$. Then $a, b, c, d$ satisfy the conditions of the lemma, so indeed there exists a balanced triangle.