Proving non-regularity via syntactic congruence classes?
You want to prove that if $\equiv_S$ has infinite index, then $L$ is not a regular language, or, equivalently that if $L$ is regular, then $\equiv_S$ has finite index. This is an immediate consequence of the following result.
Theorem. Let ${\cal A} = (Q, \Sigma, \cdot, i, F)$ be the minimal complete deterministic automaton of $L$. Then $u \equiv_S v$ if and only if, for every state $q \in Q$, $q \cdot u = q \cdot v$.
Proof. Suppose that $u \equiv_S v$ and let $q \in Q$. Since $\cal A$ is minimal, it is accessible, and there exists a word $x$ such that $i \cdot x = q$. Now, for every word $y$,
$$
xuy \in L \iff xvy \in L, 
$$
or, equivalently,
$$
1 \cdot xuy \in F \iff 1 \cdot xvy \in F
$$
that is,
$$
(q \cdot u) \cdot y \in F \iff (q \cdot v) \cdot y \in F
$$
It follows that the states $q \cdot u$ and $q \cdot v$ are equivalent, and hence equal, since $\cal A$ is minimal.
Suppose now that for every state $q \in Q$, $q \cdot u = q \cdot v$. Let $x$ and $y$ be words. Then the following equivalences hold:
\begin{align}
xuy \in L &amp;\iff 1 \cdot xuy \in F \iff ((1 \cdot x) \cdot u) \cdot y \in F \\
&amp;\iff ((1 \cdot x) \cdot v) \cdot y \in F \iff 1 \cdot xvy \in F \\
&amp;\iff xvy \in L
\end{align}
and thus $u \equiv_S v$.