Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 4 / 2D / ii / Solution

The union need not be an equivalence relation, because transitivity can mix the two relations. Take $X=\{1,2,3\}$. Let $R$ have equivalence classes $\{1,2\}$ and $\{3\}$, and $S$ have classes $\{1\}$ and $\{2,3\}$. Then $1W2$ and $2W3$, but neither component relates $1$ to $3$, so $1W3$ fails. The union of equivalence relations is reflexive and symmetric; it is precisely transitivity that may fail.