Prove $ R \cap (B \times B) $ is partial order on B Given $R$ is a partial order on $A$ and $B$ is a subset of $A$
Let $x,y,z \in B$
So $x,y,z \in A$


*

*So $(x,x) \in R $ 
Also $(x,x) \in B \times B$. Hence follows the reflexivity.

*Since $x,y,z  \in B$ and so in $A$. So let $xRy$ and $yRz$ so $xRz$ by transititvity of $R$. Also since $x,z \in B$ so $(x,z) \in B \times B$. So $xRz \in R \cap (B \times B)$

*Now let  $xRy$ and $yRx$. This implies $x=y$ because $R$ is partial order on $A$ and $x,y \in A$. Also $(x,y)$  and $(y,x) \in B \times B$. How do I proceed ? Is this even correct?
Thanks 
 A: The proof looks complete to me: In 3. you know that if $\left(x,\, y\right),\,\left(y,\, x\right)\,\in\, R\cap\left(B\times B\right)$ they must be in $R$ and thus the antisymmetry directly translates from $A$.
Detailed Proof, as I would write it:
Let $A$ be a set with partial order $R$. Let $B\subseteq A$.
Lemma: Then $R\cap\left(B\times B\right)$ is a partial order on $B$.
Proof:
1.) If $b\in B$ then $b\in A$ and thus $\left(b,b\right)\in R$. Thus $\left(b,b\right)\in R\cap\left(B\times B\right)$. Reflexivity $\checkmark$
2.) If $\left(a,b\right)\in R\cap\left(B\times B\right)$ and $\left(b,c\right)\in R\cap\left(B\times B\right)$
  then $\left(a,b\right)\in R$
  and $\left(b,c\right)\in R$.
Thus $\left(a,c\right)\in R$ as $R$ is a partial order.
Also from $\left(a,b\right)\in R\cap\left(B\times B\right)$
  and $\left(b,c\right)\in R\cap\left(B\times B\right)$
  follows that $\left\{ a,\, b,\, c\right\} \subset B$.
Thus $\left(a,c\right)\in R\cap\left(B\times B\right)$. Transitivity $\checkmark$
3.) If $\left(a,b\right)\in R\cap\left(B\times B\right)$
  and $\left(b,a\right)\in R\cap\left(B\times B\right)$
  then both $\left(a,b\right)\in R$
  and $\left(b,a\right)\in R$.
Thus $b=a$ as $R$ is a partial order. Antisymmetry $\checkmark$
qed
