Transitive Relation whose Symmetric Closure is not Transitive

Theorem
Let $S = \left\{{p, q}\right\}$, where $p$ and $q$ are distinct elements.

Let $\mathcal R = \left\{{\left({p, q}\right)}\right\}$.

Then $\mathcal R$ is transitive but its symmetric closure is not.

Proof
$\mathcal R$ is vacuously transitive because there are no elements $a, b, c \in S$ such that $a \mathrel{\mathcal R} b$ and $b \mathrel{\mathcal R} c$.

Let $\mathcal R^\leftrightarrow$ be the symmetric closure of $\mathcal R$.

Then $\mathcal R^\leftrightarrow = \mathcal R \cup \mathcal R^{-1} = \left\{{\left({p, q}\right), \left({q, p}\right)}\right\}$.

Then $p \mathrel{\mathcal R^\leftrightarrow} q$ and $q \mathrel{\mathcal R^\leftrightarrow} p$ but $p \not \mathrel{\mathcal R^\leftrightarrow} p$.

Therefore $\mathcal R^\leftrightarrow$ is not transitive.