Here's one way to build an infinite counterexample with $n=1$ for simplicity (assuming I'm identifying and fixing a typo correctly, see the comments):

We start with (say) $X_0=\{0\}, S_0=\emptyset$. Having defined $X_m, S_m$, we define $X_{m+1}, S_{m+1}$ as follows:

• To get $X_{m+1}$, we add to $X_m$ a fresh element $c_{(x,y)}$ for each pair $(x,y)\in X_m^2$ such that there is no $z\in X_m$ with $(x,z),(z,y)\in S_m$.

• To get $S_{m+1}$, we add to $S_m$ all pairs of the form $(x,c_{(x,y)})$ and $(c_{(x,y)},y)$ for $(x,y)\in X_m^2$ such that no $z\in X_m$ has $(x,z),(z,y)\in S_m$.

Now let $X=\bigcup_{m\in\mathbb{N}}X_m$, $S=\bigcup_{m\in\mathbb{N}}S_m$.

This same idea works for any $n$ and for any (infinite) cardinality.

Here's one way to build an infinite counterexample with $n=2$ for simplicity:

We start with (say) $X_0=\{0\}, S_0=\emptyset$. Having defined $X_m, S_m$, we define $X_{m+1}, S_{m+1}$ as follows:

• To get $X_{m+1}$, we add to $X_m$ a fresh element $c_{(x,y)}$ for each pair $(x,y)\in X_m^2$ such that there is no $z\in X_m$ with $(x,z),(z,y)\in S_m$.

• To get $S_{m+1}$, we add to $S_m$ all pairs of the form $(x,c_{(x,y)})$ and $(c_{(x,y)},y)$ for $(x,y)\in X_m^2$ such that no $z\in X_m$ has $(x,z),(z,y)\in S_m$.

Now let $X=\bigcup_{m\in\mathbb{N}}X_m$, $S=\bigcup_{m\in\mathbb{N}}S_m$.

This same idea works for any $n\ge 2$ and for any (infinite) cardinality.

Here's one way to build an infinite counterexample with $n=1$ for simplicity.

We start with (say) $X_0=\{0\}, S_0=\emptyset$. Having defined $X_m, S_m$, we define $X_{m+1}, S_{m+1}$ as follows:

• To get $X_{m+1}$, we add to $X_m$ a fresh element $c_{(x,y)}$ for each pair $(x,y)\in X_m^2$ such that there is no $z\in X_m$ with $(x,z),(z,y)\in S_m$.

• To get $S_{m+1}$, we add to $S_m$ all pairs of the form $(x,c_{(x,y)})$ and $(c_{(x,y)},y)$ for $(x,y)\in X_m^2$ such that no $z\in X_m$ has $(x,z),(z,y)\in S_m$.

Now let $X=\bigcup_{m\in\mathbb{N}}X_m$, $S=\bigcup_{m\in\mathbb{N}}S_m$.

Here's one way to build an infinite counterexample with $n=1$ for simplicity (assuming I'm identifying and fixing a typo correctly, see the comments):

We start with (say) $X_0=\{0\}, S_0=\emptyset$. Having defined $X_m, S_m$, we define $X_{m+1}, S_{m+1}$ as follows:

• To get $X_{m+1}$, we add to $X_m$ a fresh element $c_{(x,y)}$ for each pair $(x,y)\in X_m^2$ such that there is no $z\in X_m$ with $(x,z),(z,y)\in S_m$.

• To get $S_{m+1}$, we add to $S_m$ all pairs of the form $(x,c_{(x,y)})$ and $(c_{(x,y)},y)$ for $(x,y)\in X_m^2$ such that no $z\in X_m$ has $(x,z),(z,y)\in S_m$.

Now let $X=\bigcup_{m\in\mathbb{N}}X_m$, $S=\bigcup_{m\in\mathbb{N}}S_m$.

This same idea works for any $n$ and for any (infinite) cardinality.