Let $X,Y$ be sets so $Y\subseteq X$.
Let $R$ be equivalence relation of $X$. lets define $R|_{Y}$ to be a realtion on $Y$ so $R|_{Y}=(Y\times Y)\cap R$.
Is $R|_{B}$ reflexive? symmetric? transitive?
what I tried:
let $x\in X$. Since $R$ is reflexive $(x,x)\in R$ and the $(x,x)\in R|_{Y}$ so $R|_{Y}$ is reflexive.
let $x,y\in X$. since $R$ is symmetric - if $(x,y)\in R$ so $(y,x)\in R$ and then if $(x,y)\in R|_{Y}$ so $(y,x)\in R|_{Y}$ so $R|_{Y}$ is symmetric.
let $x,y,z\in X$. since $R$ is transitive - if $(x,y)\in R$ and $(y,z)\in R$ so $(x,z)\in R$ and then if $(x,y)\in R|_{Y}$ and $(y,z)\in R|_{Y}$ so $(x,z)\in R|_{Y}$ so $R|_{Y}$ is transitive.
edit: let $(y,y)\in Y\times Y$ so $(y,y)\in X\times X$ and since $R$ is reflexive on $X$, $\forall x\in X, xRx$ so $(y,y)\in R|_{Y}$.
Is it correct? what now?
A good intuitive way to see that $R \mid_Y$ is an equivalence relation is to convince yourself that a relation on $X$ is an equivalence relation if and only if it divides $X$ into distinct, non-empty regions. This is the essence of the idea behind equivalence relations and it's supposed to be very intuitive.
Therefore, if a relation divides $X$ into distinct non-empty regions, it will divide a non-empty subset of $X$ to distinct non-empty regions as well.
Moreover, your relation $R \mid_Y$ is nothing but the restriction of $R \subseteq X\times X$ to the elements in $Y \times Y$. If some property holds for every $(a,b) \in X \times X$ then surely it holds for every elment in $Y \times Y$ too because $Y \times Y \subseteq X \times X$.