Translation: I'm attempting the following proof, but I don't know which definition to use moving forward; I appreciate the help:
Let $R$ be an equivalence relation on $A$, and suppose that $B\subset A$. Prove that, for each $x\in B$, we have $[x]_{R|_B}=[x]_R\cap B$.
Let $x\in[x]_{R|_B}$. Then $(x,x)\in R|_B\implies x\in B\implies x\in A$ by hypothesis, which implies $(x,x)\in R\wedge x\in B$.
estoy intentando trabajar en la siguiente demostración pero estoy en este punto y no sé que definición utilizar para avanzar, agradezco las ayudas,
The following picture shows my attempt.
This is essentially just a matter of "unwrapping" the definitions; fix $x\in B$. Now, recall that $R|_{B}$ is the intersection of $R$ with $B\times B$, and that $[x]_{R|_B}$ is the set of all $y$ such that $(x,y)\in R|_B$. Thus we have: \begin{align} y\in[x]_{R|_B}&\iff (x,y)\in R|_B=R\cap (B\times B) \\ &\iff (x,y)\in R\text{ and }(x,y)\in B\times B \\ &\iff y\in[x]_R\text{ and }y\in B \\ &\iff y\in [x]_R\cap B, \end{align} which shows the desired equality.