No graph on exactly three vertices is prime

By [F3] the graph has at most three edges, so $\mathrm{\mid }E(G)\mathrm{\mid }$ is $0$, $1$, $2$ or $3$, and any two-element $M\subseteq V(G)$ has $\mathrm{\mid }M\mathrm{\mid }=2=\mathrm{\mid }V(G)\mathrm{\mid }-1$, hence is nontrivial once it is a module.

First case: $\mathrm{\mid }E(G)\mathrm{\mid }=0$. Take $M=\{x,y\}$; the only vertex outside is $z$, and it is adjacent to neither, so $(\{z\},M)$ is anticomplete.

Second case: $\mathrm{\mid }E(G)\mathrm{\mid }=3$. Take $M=\{x,y\}$; the only vertex outside is $z$, and by [F3] both $\{x,z\}$ and $\{y,z\}$ are edges, so $(\{z\},M)$ is complete.

Third case: $\mathrm{\mid }E(G)\mathrm{\mid }=1$, say the single edge is $\{p,q\}$ and $r$ is the remaining vertex. Take $M=\{p,q\}$; neither $\{p,r\}$ nor $\{q,r\}$ is an edge, since there is only one edge and it is $\{p,q\}$, so $(\{r\},M)$ is anticomplete.

Fourth case: $\mathrm{\mid }E(G)\mathrm{\mid }=2$. Each of the three possible edges listed in [F3] meets each of the other two, so the two edges of $G$ share a vertex $q$; write them as $\{p,q\}$ and $\{q,r\}$ with $\{p,q,r\}=V(G)$. Take $M=\{p,r\}$; the only vertex outside is $q$, which is adjacent to both, so $(\{q\},M)$ is complete.

The four cases cover every value of $\mathrm{\mid }E(G)\mathrm{\mid }$ allowed by step 1.1, and in each of them the exhibited two-element set $M$ has $(\{v\},M)$ pure for the single vertex $v$ outside it, so $M$ is a module.

That module is nontrivial by step 1.1, so $G$ is not prime.