Fake proof, symmetric and transitive relation is already reflexive

Question

Let $R$ be a symmetric, transitive relation. If $(x, y) \in R$ then the symmetric property implies that $(y, x) \in R$. Using the the transitive property upon $(x, y)$ and $(y, x)$ we can conclude $(x, x) \in R$. Is this fair logic or is it flawed?

Answer 1

You may conclude that for all $x$, if $(x, y)\in R$ for some $y$ then $(x,x)\in R$. That's not the same that proving that $(x,x) \in R$ for all $x$.