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The printed assertion is false when maximal means largest in the natural-number labelling. Both the original PDF and the TeX have this wording. The specified operation gives the complete graph, not the Rado graph.

Fix any two distinct vertices $u,v\in \mathbb{N}$, and choose $M\ge \max \{u,v\}$. Apply the hypergraph extension property of $G_{\mathrm{univ}}^{(3)}$ to $S=\{1,\dots ,M\}$, prescribing that the pair $\{u,v\}$ extend to a hyperedge; prescribe all other pairs in $S$ arbitrarily. The resulting witness $w\notin S$ has $w>M$, so $\{u,v,w\}$ is a hyperedge whose largest vertex is $w$. Deleting that vertex produces the edge $\{u,v\}$ in $G_0$.

Since $u,v$ were arbitrary,The Rado graph has nonedges: its hypergraph extension property supplies a new vertex nonadjacent to any specified existing vertex. Therefore $K_{\mathbb{N}}$ and $G_{\mathrm{univ}}^{(2)}$ cannot be isomorphic graphs. This argument holds for every representative of the universal three-uniform structure labelled by $\mathbb{N}$, so changing its labelling does not repair the assertion. Equivalently, in an independently sampled representative each pair has infinitely many larger vertices that complete it to a triple, almost surely.

A valid related construction is to fix a vertex $a$ and take its link hypergraph: make $\{u,v\}$ an edge exactly when $\{a,u,v\}$ is a hyperedge, on the remaining vertex set. This is indeed the Rado graph. To prove it, take any disjoint finite sets $U,W$ of vertices other than $a$, put $S=\{a\}\cup U\cup W$, and prescribe triples through the pairs $\{a,u\}$ for $u\in U$ and no triples through $\{a,w\}$ for $w\in W$. Complete all other pair prescriptions arbitrarily. The hypergraph extension property produces a new vertex adjacent in the link to all of $U$ and none of $W$. Uniqueness from part (i), for $r=2$, now proves the claimed hypergraph isomorphism for this corrected operation.