Let $S$ be a equivalence relation on $A$, let $B$ be a subset of $A$ and suppose that $T$ equivalence relation on $B$. Defining $R=S \cup T$. Its given that there is exist $x\in A$ such that $B$ is equivalence class of $x$ to $S$. Q: Is R equivalence relation?
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The relation $xRy$ is equvalent to "$xSy$ or $xTy$". Since $B$ is an equivalence class w.r.t $S$, the case $xTy$ (which only applies when $x,y \in B$) automatically implies $xSy$ in any case in which $x,y$ happen to both be in $B$. This all means that actually you have $T$ is a subset of $S$, so that $R=S \cup T=S$. Thus since $S$ was assumed to be an equivalence relation, the relation $R$ is also because $R=S$.
Here's a concrete example: Let $A$ be the set of integers, and define $xSy$ to mean that $x-y$ is even. Let $B$ be the set of even integers, and for $x,y \in B$ define $xTy$ to mean that $(x-y)/2$ is even. Then we have that $B$ is the equivalence class of $0 \in A$ under the relation $S$. In this example you can see that $R=S\cup T=S$.